IB Applications and Interpretation SL · Arithmetic and geometric sequences · End of topic

IB exam questions for topic 2

107 exercises with solutions · By Álvaro López

Past IB exam questions grouped by type and, within each type, from easiest to hardest. Try each one before opening the solution.

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Paper

Solutions

General term and properties of sequences

Exercise 1 BasicIB · P1 TZ0 November 2005

a) The first term of an arithmetic sequence is $- 16$ and the eleventh term is 39.

Calculate the common difference of the sequence.

b) The third term of a geometric sequence is 12 and the fifth term is $\frac{16}{3}$ .

All the terms of the sequence are positive. Calculate the common ratio of the sequence.

Show answer

Answer: a) $d = 5.5$ ; b) $r = \frac{2}{3}$

Show working and answer

a) Arithmetic sequence with $u_{1} = - 16$ and $u_{11} = 39$ .

$u_{11} = u_{1} + 10 d$ , that is, $39 = - 16 + 10 d$ .
$10 d = 55$ , so $d = 5.5$ .

b) Geometric sequence with $u_{3} = 12$ and $u_{5} = \frac{16}{3}$ .

$u_{5} = u_{3} r^{2}$ , since there are two steps between the third and the fifth term.
$\frac{16}{3} = 12 r^{2}$ , so $r^{2} = \frac{16}{36} = \frac{4}{9}$ .
Hence $r = \pm \frac{2}{3}$ ; as all the terms are positive, $r > 0$ and so $r = \frac{2}{3}$ .

Answer: a) $d = 5.5$ ; b) $r = \frac{2}{3}$

Exercise 2 BasicIB · P1 TZ2 May 2015

Only one of the following four sequences is arithmetic and only one of them is geometric.

$a_{n} = 1; 2; 3; 5; \dots$

$b_{n} = 1; \frac{3}{2}; \frac{9}{4}; \frac{27}{8}; \dots$

$c_{n} = 1; \frac{1}{2}; \frac{1}{3}; \frac{1}{4}; \dots$

$d_{n} = 1; 0.95; 0.90; 0.85; \dots$

a) State which sequence is

(i) arithmetic;

(ii) geometric.

b) For another geometric sequence $e_{n} = - 6; - 3; - \frac{3}{2}; - \frac{3}{4}; \dots$

(i) write down the common ratio;

(ii) find the exact value of the tenth term.

Give your answer as a fraction.

Show answer

Answer: a) (i) $d_{n}$ ; (ii) $b_{n}$ . b) (i) $r = \frac{1}{2}$ ; (ii) $e_{10} = - \frac{3}{256}$ .

Show working and answer

a) A sequence is arithmetic if the difference between consecutive terms is constant, and geometric if the quotient of consecutive terms is constant.

$a_{n}$ : $2 - 1 = 1$ and $3 - 2 = 1$ , but $5 - 3 = 2$ . It is not arithmetic. Quotients: $2, 1.5, 1.67$ , so it is not geometric either.
$b_{n}$ : $\frac{3/2}{1} = \frac{3}{2}$ , $\frac{9/4}{3/2} = \frac{3}{2}$ , $\frac{27/8}{9/4} = \frac{3}{2}$ . Constant quotient: geometric.
$c_{n}$ : differences $- \frac{1}{2}$ , $- \frac{1}{6}$ , not constant; quotients $\frac{1}{2}, \frac{2}{3}$ , not constant. Neither.
$d_{n}$ : $0.95 - 1 = - 0.05$ , $0.90 - 0.95 = - 0.05$ , $0.85 - 0.90 = - 0.05$ . Constant difference: arithmetic.

(i) arithmetic: $d_{n}$ . (ii) geometric: $b_{n}$ .

(i) Common ratio: $r = \frac{- 3}{- 6} = \frac{1}{2}$ . Check: $\frac{- 3/2}{- 3} = \frac{1}{2}$ .
(ii) General term $e_{n} = u_{1} r^{n - 1}$ , with $u_{1} = - 6$ : $e_{10} = - 6 (\frac{1}{2})^{9} = - \frac{6}{512} = - \frac{3}{256}$

Answer: a) (i) $d_{n}$ ; (ii) $b_{n}$ . b) (i) $r = \frac{1}{2}$ ; (ii) $e_{10} = - \frac{3}{256}$ .

Exercise 3 BasicIB · P1 TZ0 November 2018

The following table shows the first five terms of three sequences: $u_{n}$ , $v_{n}$ and $w_{n}$ .

$n$	1	2	3	4	5
$u_{n}$	10	20	40	80	160
$v_{n}$	10	20	30	60	100
$w_{n}$	10	20	30	40	50

a) State which of these sequences is

(i) arithmetic;

(ii) geometric.

b) Find the exact value of the 11th term of the geometric sequence.

c) Find the sum of the first 20 terms of the arithmetic sequence.

Show answer

Answer: a) (i) $w_{n}$ ; (ii) $u_{n}$ . b) $u_{11} = 10240$ . c) $S_{20} = 2100$ .

Show working and answer

$u_{n}$ : quotients $\frac{20}{10} = \frac{40}{20} = \frac{80}{40} = \frac{160}{80} = 2$ . It is geometric with $r = 2$ .
$v_{n}$ : differences $10, 10, 30, 40$ and quotients $2, 1.5, 2$ ; it is neither.
$w_{n}$ : differences all equal to $10$ . It is arithmetic with $d = 10$ .

(i) arithmetic: $w_{n}$ . (ii) geometric: $u_{n}$ .

b) With $u_{1} = 10$ and $r = 2$ : $u_{n} = 10 \cdot 2^{n - 1}$ , so $u_{11} = 10 \cdot 2^{10} = 10 \cdot 1024 = 10240$

c) With $w_{1} = 10$ and $d = 10$ we use $S_{n} = \frac{n}{2} (2 w_{1} + (n - 1) d)$ : $S_{20} = \frac{20}{2} (2 \cdot 10 + 19 \cdot 10) = 10 \cdot 210 = 2100$

Answer: a) (i) $w_{n}$ ; (ii) $u_{n}$ . b) $u_{11} = 10240$ . c) $S_{20} = 2100$ .

Exercise 4 MediumIB · P2 TZ1 May 2014

Consider the sequence $u_{1}, u_{2}, u_{3}, \dots\dots, u_{n}, \dots\dots$ where $u_{1} = 600, u_{2} = 617, u_{3} = 634, u_{4} = 651$ .

The sequence continues in the same way.

a) Find the value of $u_{20}$ .

b) Find the sum of the first 10 terms of the sequence.

Now consider the sequence $v_{1}, v_{2}, v_{3}, \dots\dots, v_{n}, \dots\dots$ where $v_{1} = 3, v_{2} = 6, v_{3} = 12, v_{4} = 24$

This sequence continues in the same way.

c) Find the exact value of $v_{10}$ .

d) Find the sum of the first 8 terms of this sequence.

$k$ is the smallest value of $n$ for which $v_{n}$ is greater than $u_{n}$

e) Calculate the value of $k$ .

Show answer

Answer: a) $u_{20} = 923$ . b) $S_{10} = 6765$ . c) $v_{10} = 1536$ . d) $S_{8} = 765$ . e) $k = 9$ .

Show working and answer

First sequence: the differences are $617 - 600 = 17$ , $634 - 617 = 17$ , $651 - 634 = 17$ ; it is arithmetic with $u_{1} = 600$ and $d = 17$ .

a) $u_{20} = u_{1} + 19 d = 600 + 19 \cdot 17 = 600 + 323 = 923$ .

b) $S_{10} = \frac{10}{2} (2 \cdot 600 + 9 \cdot 17) = 5 (1200 + 153) = 5 \cdot 1353 = 6765$ .

Second sequence: the quotients are $\frac{6}{3} = \frac{12}{6} = \frac{24}{12} = 2$ ; it is geometric with $v_{1} = 3$ and $r = 2$ .

c) $v_{10} = 3 \cdot 2^{9} = 3 \cdot 512 = 1536$ .

d) $S_{8} = \frac{3 ( 2 ^{8} - 1 )}{2 - 1} = 3 \cdot 255 = 765$ .

We look for the smallest $n$ with $v_{n} > u_{n}$ , that is $3 \cdot 2^{n - 1} > 600 + 17 (n - 1)$ . This can be solved using the GDC by comparing the two sequences in a table.
$n = 8$ : $v_{8} = 3 \cdot 128 = 384$ and $u_{8} = 600 + 7 \cdot 17 = 719$ . Still $v_{8} < u_{8}$ .
$n = 9$ : $v_{9} = 3 \cdot 256 = 768$ and $u_{9} = 600 + 8 \cdot 17 = 736$ . Now $v_{9} > u_{9}$ .
As $v_{n}$ grows faster and the condition has not been met before, $k = 9$ .

Answer: a) $u_{20} = 923$ . b) $S_{10} = 6765$ . c) $v_{10} = 1536$ . d) $S_{8} = 765$ . e) $k = 9$ .

Exercise 5 MediumIB · P1 TZ1 May 2008

The first term of an arithmetic sequence is 0 and the common difference is 12.

a) Find the value of the 96th term of the sequence.

The first term of a geometric sequence is 6. The sixth term of the geometric sequence is equal to the 17th term of the arithmetic sequence given above.

b) Write down an equation using this information.

c) Calculate the common ratio of the geometric sequence.

Show answer

Answer: $u_{96} = 1140$ ; equation $6 r^{5} = 192$ ; $r = 2$ .

Show working and answer

$u_{n} = u_{1} + (n - 1) d$ with $u_{1} = 0$ and $d = 12$ .
$u_{96} = 0 + 95 \cdot 12 = 1140$ .

Answer a): $u_{96} = 1140$ .

The 17th term of the arithmetic sequence is $u_{17} = 16 \cdot 12 = 192$ .
In the geometric sequence $v_{n} = v_{1} r^{n - 1}$ , with $v_{1} = 6$ , the sixth term is $6 r^{5}$ .
Equating: $6 r^{5} = 192$ .

Answer b): $6 r^{5} = 192$ .

Solving for $r$ : $r^{5} = \frac{192}{6} = 32$ .
$r = 532 = 2$ (because $2^{5} = 32$ ).

Answer c): $r = 2$ .

Answer: $u_{96} = 1140$ ; equation $6 r^{5} = 192$ ; $r = 2$ .

Exercise 6 MediumIB · P1 TZ1 May 2015

The second term of an arithmetic sequence is 30. The fifth term is 90.

a) Calculate

(i) the common difference of the sequence;

(ii) the first term of the sequence.

The first, second and fifth terms of this arithmetic sequence are the first three terms of a geometric sequence.

b) Calculate the seventh term of the geometric sequence.

Show answer

Answer: $d = 20$ ; $u_{1} = 10$ ; $v_{7} = 7290$ .

Show working and answer

a) With $u_{2} = u_{1} + d = 30$ and $u_{5} = u_{1} + 4 d = 90$ :

Subtracting: $3 d = 60$ , so $d = 20$ .
$u_{1} = 30 - 20 = 10$ .

Answer a): (i) $d = 20$ ; (ii) $u_{1} = 10$ .

b) 1. The first three terms of the geometric sequence are $10, 30, 90$ . 2. Ratio: $r = \frac{30}{10} = 3$ (and $\frac{90}{30} = 3$ , which is consistent). 3. Seventh term: $v_{7} = 10 \cdot 3^{6} = 10 \cdot 729 = 7290$ .

Answer b): $v_{7} = 7290$ .

Answer: $d = 20$ ; $u_{1} = 10$ ; $v_{7} = 7290$ .

Exercise 7 MediumIB · P2 TZ0 November 2003

Ana and Juan are going swimming at the pool.

Both of them swim the first length of the pool in 2 minutes.

Each length Juan swims, he takes 6 seconds longer than on the previous length.

Each length Ana swims, she takes 1.05 times as long as on the previous length.

a) (i) Calculate the time Juan takes to swim the third length.

(ii) Verify that Ana takes 2.205 minutes to swim the third length.

b) Calculate the time Ana takes to swim a total of 10 lengths of the pool.

Show answer

Answer: Juan: 2 min 12 s; Ana: 2.205 min; Ana's total $\approx 25.2$ min.

Show working and answer

(i) Juan's times form an arithmetic sequence with $u_{1} = 2$ min and $d = 6$ s $= 0.1$ min.

$u_{3} = 2 + 2 \cdot 0.1 = 2.2$ min.
$2.2$ min is $2$ min and $12$ s.

(ii) Ana's times form a geometric sequence with $u_{1} = 2$ and $r = 1.05$ .

$u_{3} = 2 \cdot 1.0 5^{2} = 2 \cdot 1.1025 = 2.205$ min.

Proved.

Answer a): (i) Juan takes $2$ min $12$ s on the third length; (ii) Ana takes $2.205$ min.

b) 1. Geometric sum: $S_{10} = \frac{2 ( 1.0 5 ^{10} - 1 )}{1.05 - 1}$ . 2. $S_{10} = \frac{2 \cdot 0.628895}{0.05} \approx 25.16$ min.

Answer b): Ana takes approximately $25.2$ minutes.

Answer: Juan: 2 min 12 s; Ana: 2.205 min; Ana's total $\approx 25.2$ min.

Exercise 8 HardIB · P2 TZ1 May 2012

Part A

The Green Park Amphitheatre was built in the shape of a horseshoe and has 20 rows.

The number of seats in each row increases by a fixed amount, $d$ , compared with the number of seats in the previous row. The number of seats in the sixth row, $u_{6}$ , is 100, and the number of seats in the tenth row, $u_{10}$ , is 124.

$u_{1}$ represents the number of seats in the first row.

a) (i) Write down an equation for $u_{6}$ in terms of $d$ and $u_{1}$ .

(ii) Write down an equation for $u_{10}$ in terms of $d$ and $u_{1}$ .

b) Write down the value of

(i) $d$ ;

(ii) $u_{1}$ .

c) Find the total number of seats in the amphitheatre.

Some years later, a second tier was added to increase the capacity of the amphitheatre by another 1,600 seats. Each row has four more seats than the row before it.

The first row in this tier has 70 seats.

d) Find the number of rows in the second tier of the amphitheatre.

Part B

Frank is in the amphitheatre and receives a text message at 12:00. Five minutes later he forwards the text message to three people. Five minutes later, those three people forward the text message to three new people. Suppose this pattern continues, and each time the text message is sent to people who have not received it before.

The number of new people who receive the text message forms a geometric sequence.

$1, 3, \dots$

a) Write down the next two terms of this geometric sequence.

b) Write down the common ratio of this geometric sequence.

c) Calculate the number of people who will receive the text message at 12:30.

d) Calculate the total number of people who will have received the text message by 12:30.

e) Calculate the exact time at which a total of 29,524 people will have received the text message.

Show answer

Answer: A: a) (i) $u_{1} + 5 d = 100$ ; (ii) $u_{1} + 9 d = 124$ ; b) $d = 6$ , $u_{1} = 70$ ; c) $2540$ seats; d) $16$ rows. B: a) $9, 27$ ; b) $r = 3$ ; c) $729$ people; d) $1093$ people; e) at $12 : 45$ .

Show working and answer

Part A

a) The general term is $u_{n} = u_{1} + (n - 1) d$ .

(i) $u_{6} = u_{1} + 5 d$ , and it equals $100$ : $u_{1} + 5 d = 100$ .

(ii) $u_{10} = u_{1} + 9 d$ , and it equals $124$ : $u_{1} + 9 d = 124$ .

b) We subtract the two equations: $(u_{1} + 9 d) - (u_{1} + 5 d) = 124 - 100$ , that is, $4 d = 24$ .

(i) $d = 6$ .

(ii) From $u_{1} + 5 \cdot 6 = 100$ we obtain $u_{1} = 70$ .

c) There are 20 rows. We use $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ :

$S_{20} = \frac{20}{2} (2 \cdot 70 + 19 \cdot 6) = 10 (140 + 114) = 10 \cdot 254 = 2540$ .

The amphitheatre has $2540$ seats.

d) In the second tier $u_{1} = 70$ , $d = 4$ and the sum must be $1600$ :

$\frac{n}{2} (2 \cdot 70 + (n - 1) \cdot 4) = 1600$
$n (70 + 2 (n - 1)) = 1600 \Rightarrow n (2 n + 68) = 1600 \Rightarrow 2 n^{2} + 68 n - 1600 = 0 \Rightarrow n^{2} + 34 n - 800 = 0$ .
Factorising, $(n - 16) (n + 50) = 0$ , so $n = 16$ (the solution $n = - 50$ is not valid).

The second tier has $16$ rows.

Part B

a) Each term is obtained by multiplying the previous one by 3: $1, 3, 9, 27, \dots$ The next two are $9$ and $27$ .

b) $r = \frac{3}{1} = 3$ .

c) Messages are sent every 5 minutes: 12:00 corresponds to term 1, 12:05 to term 2, and so on. 12:30 is $30/5 = 6$ intervals later, that is, term $n = 7$ :

$u_{7} = 1 \cdot 3^{6} = 729$ .

d) Total up to 12:30 = sum of the first 7 terms:

$S_{7} = \frac{1 \cdot ( 3 ^{7} - 1 )}{3 - 1} = \frac{2187 - 1}{2} = 1093$ .

e) We look for $n$ with $S_{n} = \frac{3 ^{n} - 1}{2} = 29524$ :

$3^{n} - 1 = 59048 \Rightarrow 3^{n} = 59049$ .
As $59049 = 3^{10}$ , $n = 10$ .
Term 10 corresponds to 9 intervals of 5 minutes after 12:00, that is, $45$ minutes: 12:45.

Exercise 9 HardIB · P2 TZ2 May 2011

Part A

A geometric sequence has 1024 as its first term and 128 as its fourth term.

a) Show that the common ratio is $\frac{1}{2}$ .

b) Find the value of the eleventh term.

c) Find the sum of the first eight terms.

d) Find the number of terms in the sequence for which the sum of the first terms exceeds 2,047.968.

Part B

Consider the arithmetic sequence $1, 4, 7, 10, 13, \dots$

a) Find the value of the eleventh term.

b) The sum of the first $n$ terms of this sequence is $\frac{n}{2} (3 n - 1)$ .

(i) Find the sum of the first 100 terms of this arithmetic sequence.

(ii) The sum of the first $n$ terms is 477.

a) Show that $3 n^{2} - n - 954 = 0$ .

b) Using your GDC or otherwise, find the number of terms, $n$ .

Show answer

Answer: A: b) $u_{11} = 1$ ; c) $S_{8} = 2040$ ; d) $16$ terms. B: a) $31$ ; b) (i) $14950$ ; (ii) $n = 18$ .

Show working and answer

Part A

a) $u_{4} = u_{1} r^{3}$ , so $128 = 1024 r^{3}$ and $r^{3} = \frac{128}{1024} = \frac{1}{8}$ . Therefore $r = 3 \frac{1}{8} = \frac{1}{2}$ . Proved.

b) $u_{11} = 1024 (\frac{1}{2})^{10} = \frac{1024}{1024} = 1$ .

c) $S_{8} = \frac{1024 ( 1 - ( \frac{1}{2} ) ^{8} )}{1 - \frac{1}{2}} = 2048 (1 - \frac{1}{256}) = 2048 - 8 = 2040$ .

d) $S_{n} = 2048 (1 - (\frac{1}{2})^{n}) > 2047.968$ :

$2048 (\frac{1}{2})^{n} < 2048 - 2047.968 = 0.032$ .
$(\frac{1}{2})^{n} < \frac{0.032}{2048} = \frac{1}{64 000}$ , that is, $2^{n} > 64000$ .
As $2^{15} = 32768 < 64000$ and $2^{16} = 65536 > 64000$ , the smallest $n$ is $16$ (this can also be checked with a table on the GDC).

The number of terms is $16$ .

Part B

a) $d = 3$ and $u_{1} = 1$ : $u_{11} = 1 + 10 \cdot 3 = 31$ .

b) (i) $S_{100} = \frac{100}{2} (3 \cdot 100 - 1) = 50 \cdot 299 = 14950$ .

(ii) (a) We equate $\frac{n}{2} (3 n - 1) = 477$ :

$n (3 n - 1) = 954$
$3 n^{2} - n = 954$
$3 n^{2} - n - 954 = 0$ . Proved.

b) We solve the equation using the GDC (or the formula): $n = \frac{1 \pm 1 + 4 \cdot 3 \cdot 954}{6} = \frac{1 \pm 107}{6}$ , which gives $n = 18$ or $n = - \frac{53}{3}$ . As $n$ must be a positive integer, $n = 18$ .

Answer: A: b) $u_{11} = 1$ ; c) $S_{8} = 2040$ ; d) $16$ terms. B: a) $31$ ; b) (i) $14950$ ; (ii) $n = 18$ .

Exercise 10 HardIB · P2 TZ2 May 2019

On the first day Kiri is in hospital, she is given $u_{1}$ milligrams (mg) of a certain medication.

The amount of medication Kiri is given increases each day, always by the same amount ( $d$ ). On the seventh day she is given 21 mg of medication, and on the eleventh day she is given 29 mg.

a) Write down an equation, in terms of $u_{1}$ and $d$ , that represents the amount of medication she is given on

(i) the seventh day;

(ii) the eleventh day.

b) Write down the value of $d$ and the value of $u_{1}$ .

Kiri takes this medication for 30 days.

c) Calculate the total amount of the medication (in mg) she will have taken.

Ted is also in hospital, and on the first day he is given an injection of 20 mg of antibiotic. The amount of antibiotic Ted is given decreases by 50% each day.

That is, on the second day Ted is given an injection of 10 mg of antibiotic, on the third day he is given 5 mg, and so on.

d) (i) Find the amount of antibiotic (in mg) Ted is given on the fifth day.

(ii) The daily amount of antibiotic Ted is given will be less than 0.06 mg for the first time on the $k$ th day. Find the value of $k$ .

(iii) Hence, find the total amount of antibiotic (in mg) Ted will have been given during those first $k$ days.

Show answer

Answer: $d = 2$ , $u_{1} = 9$ ; Kiri takes $1140$ mg; Ted: $1.25$ mg on the fifth day, $k = 10$ , and $\approx 40.0$ mg in total.

Show working and answer

a) The daily increases are constant, so $u_{n} = u_{1} + (n - 1) d$ .

(i) Seventh day: $u_{1} + 6 d = 21$ .

(ii) Eleventh day: $u_{1} + 10 d = 29$ .

Answer a): $u_{1} + 6 d = 21$ and $u_{1} + 10 d = 29$ .

b) 1. Subtracting the two equations: $4 d = 8$ , so $d = 2$ . 2. $u_{1} + 12 = 21$ , so $u_{1} = 9$ .

Answer b): $d = 2$ mg; $u_{1} = 9$ mg.

c) 1. $S_{30} = \frac{30}{2} (2 \cdot 9 + 29 \cdot 2)$ . 2. $S_{30} = 15 (18 + 58) = 15 \cdot 76 = 1140$ mg.

Answer c): $1140$ mg.

d) Ted's doses form a geometric sequence with $v_{1} = 20$ and $r = 0.5$ .

(i) $v_{5} = 20 \cdot 0. 5^{4} = 20 \cdot 0.0625 = 1.25$ mg.

(ii) 1. We require $20 \cdot 0. 5^{k - 1} < 0.06$ , that is, $0. 5^{k - 1} < 0.003$ . 2. Using the GDC (a table of values or solving the inequality) or logarithms: $k - 1 > \frac{ln 0.003}{ln 0.5} \approx 8.38$ . 3. The smallest integer is $k - 1 = 9$ , so $k = 10$ . Check: $v_{9} = 0.078 > 0.06$ and $v_{10} = 0.039 < 0.06$ .

(iii) 1. $S_{10} = \frac{20 ( 1 - 0. 5 ^{10} )}{1 - 0.5} = 40 (1 - \frac{1}{1024})$ . 2. $S_{10} \approx 39.96 \approx 40.0$ mg.

Answer d): (i) $1.25$ mg; (ii) $k = 10$ (tenth day); (iii) $\approx 40.0$ mg.

Answer: $d = 2$ , $u_{1} = 9$ ; Kiri takes $1140$ mg; Ted: $1.25$ mg on the fifth day, $k = 10$ , and $\approx 40.0$ mg in total.

Exercise 11 HardIB · P2 TZ2 May 2016

Prachi is on holiday in the United States. She is visiting the Grand Canyon.

When she reaches the highest point of the canyon, she drops a coin over the edge of a cliff.

As it falls, the coin travels a distance of 5 metres in the first second, 15 metres in the next second, 25 metres in the third second, and it keeps falling in this way.

The distances the coin travels in each second form an arithmetic sequence.

a) (i) Write down the common difference, $d$ , of this arithmetic sequence.

(ii) Write down the distance the coin travels in the fourth second.

b) Calculate the distance the coin travels in the 15th second.

c) Calculate the total distance the coin travels in the first 15 seconds of its fall.

Give your answer in kilometres.

Prachi drops the coin from a height of 1,800 metres above the ground.

d) Calculate the time, rounded to the nearest whole number of seconds, that it takes the coin to reach the ground.

Prachi goes to a nearby tourist office. It opened at the beginning of 2015, and in its first year it already had 17,000 visitors. The number of people visiting this tourist office is expected to increase by 10% each year.

e) Calculate how many people are expected to visit the tourist office in 2016.

f) Calculate the total number of people expected to have visited the office during the first 10 years after it opened.

Show answer

Answer: a) (i) $d = 10$ m; (ii) 35 m. b) 145 m. c) $1.125$ km ( $\approx 1.13$ km). d) 19 s. e) 18,700 people. f) $\approx 270936$ people ( $\approx 271000$ ).

Show working and answer

The distances $5, 15, 25, \dots$ form an arithmetic sequence with $u_{1} = 5$ .

(i) The common difference is the difference between two consecutive terms: $d = 15 - 5 = 10$ m.
(ii) Fourth second: $u_{4} = u_{3} + d = 25 + 10 = 35$ m.

The general term formula is $u_{n} = u_{1} + (n - 1) d$ .
For $n = 15$ : $u_{15} = 5 + 14 \cdot 10 = 145$ m.

Sum of the first $n$ terms: $S_{n} = \frac{n}{2} (u_{1} + u_{n})$ .
$S_{15} = \frac{15}{2} (5 + 145) = 1125$ m.
Converting to kilometres by dividing by 1000: $1.125$ km $\approx 1.13$ km.

The general sum is $S_{n} = \frac{n}{2} [2 \cdot 5 + (n - 1) \cdot 10] = 5 n^{2}$ .
We impose $S_{n} = 1800$ : $5 n^{2} = 1800 \Rightarrow n^{2} = 360 \Rightarrow n \approx 18.97$ .
Rounding to the nearest integer: $n = 19$ seconds.

e) Each year the number of visitors is multiplied by $1.10$ (a geometric sequence with $v_{1} = 17000$ in 2015).

For 2016: $v_{2} = 17000 \cdot 1.1 = 18700$ people.

Sum of the first $n$ terms of a geometric sequence: $S_{n} = \frac{v _{1} ( r ^{n} - 1 )}{r - 1}$ , with $r = 1.1$ .
$S_{10} = \frac{17000 ( 1. 1 ^{10} - 1 )}{0.1} \approx 270936$ .

Answer: a) (i) $d = 10$ m; (ii) 35 m. b) 145 m. c) $1.125$ km ( $\approx 1.13$ km). d) 19 s. e) 18,700 people. f) $\approx 270936$ people ( $\approx 271000$ ).

Exercise 12 HardIB · P2 TZ2 May 2013

On Monday Paco goes training at an athletics track. He runs the first lap of the track in 120 seconds.

From then on, each lap Paco runs takes 10 seconds longer than the previous lap.

a) Find the time, in seconds, Paco takes to run the fifth lap.

Paco takes 260 seconds to run the last lap.

b) Find how many laps Paco ran on Monday.

c) Find the total time, in minutes, that Paco was running on Monday.

On Wednesday Paco takes Lola to training. They both run the first lap of the track in 120 seconds. Each lap Lola runs takes 1.06 times as long as the previous lap.

d) Find the time, in seconds, Lola takes to run the third lap.

e) Find the total time, in seconds, Lola takes to run the first four laps.

As before, each lap Paco runs takes 10 seconds longer than the lap before. After a certain number of laps, Paco runs each lap in less time than Lola.

f) Find from which lap this happens.

Show answer

Answer: a) 160 s. b) 15 laps. c) 47.5 min. d) $\approx 134.83$ s ( $\approx 135$ s). e) $\approx 524.95$ s ( $\approx 525$ s). f) from lap 13.

Show working and answer

a) Paco's times form an arithmetic sequence with $u_{1} = 120$ and $d = 10$ .

$u_{5} = 120 + 4 \cdot 10 = 160$ s.

We set $u_{n} = 260$ : $120 + (n - 1) \cdot 10 = 260$ .
$(n - 1) \cdot 10 = 140 \Rightarrow n - 1 = 14 \Rightarrow n = 15$ laps.

Arithmetic sum: $S_{15} = \frac{15}{2} (120 + 260) = 2850$ s.
In minutes: $2850/60 = 47.5$ min.

d) Lola's times form a geometric sequence with $v_{1} = 120$ and $r = 1.06$ .

$v_{n} = v_{1} r^{n - 1}$ .
$v_{3} = 120 \cdot 1.0 6^{2} = 134.832 \approx 134.83$ s ( $\approx 135$ s).

$S_{4} = \frac{120 ( 1.0 6 ^{4} - 1 )}{1.06 - 1}$ .
$S_{4} \approx 524.95$ s ( $\approx 525$ s).

Paco's time on lap $n$ is $u_{n} = 120 + 10 (n - 1) = 110 + 10 n$ . Lola's time is $v_{n} = 120 \cdot 1.0 6^{n - 1}$ .
We look for the smallest $n$ with $110 + 10 n < 120 \cdot 1.0 6^{n - 1}$ . This can be done with a table on the GDC, or by solving the inequality using the two graphs.
Checking values: $n = 12$ : Paco $230$ , Lola $\approx 227.8$ (Lola is still faster). $n = 13$ : Paco $240$ , Lola $\approx 241.5$ (so Paco is now faster).

Answer: a) 160 s. b) 15 laps. c) 47.5 min. d) $\approx 134.83$ s ( $\approx 135$ s). e) $\approx 524.95$ s ( $\approx 525$ s). f) from lap 13.

Arithmetic sequences: term and sum

Exercise 13 BasicIB · P1 TZ0 May 2005

The first five terms of an arithmetic sequence are shown below.

$2, 6, 10, 14, 18$

a) Write down the sixth term of this sequence.

b) Calculate the 200th term.

c) Calculate the sum of the first 90 terms of the sequence.

Show answer

Answer: a) $u_{6} = 22$ ; b) $u_{200} = 798$ ; c) $S_{90} = 16200$

Show working and answer

The difference between consecutive terms is $6 - 2 = 4$ , so $u_{1} = 2$ and $d = 4$ .

a) $u_{6} = u_{5} + d = 18 + 4 = 22$ .

b) $u_{200} = u_{1} + 199 d = 2 + 199 \cdot 4 = 2 + 796 = 798$ .

c) $S_{90} = \frac{90}{2} (2 \cdot 2 + 89 \cdot 4) = 45 (4 + 356) = 45 \cdot 360 = 16200$ .

Answer: a) $u_{6} = 22$ ; b) $u_{200} = 798$ ; c) $S_{90} = 16200$

Exercise 14 BasicIB · P1 TZ0 November 2001

The first four terms of an arithmetic sequence are shown below.

$1, 5, 9, 13, \dots\dots$

a) Write down the $n$ th term of the sequence.

b) Calculate the 100th term of the sequence.

c) Find the sum of the first 100 terms of the sequence.

Show answer

Answer: a) $u_{n} = 4 n - 3$ ; b) $u_{100} = 397$ ; c) $S_{100} = 19900$

Show working and answer

The common difference is $d = 5 - 1 = 4$ and $u_{1} = 1$ .

a) $u_{n} = u_{1} + (n - 1) d = 1 + 4 (n - 1) = 4 n - 3$ .

b) $u_{100} = 4 \cdot 100 - 3 = 397$ .

c) $S_{100} = \frac{100}{2} (u_{1} + u_{100}) = 50 (1 + 397) = 50 \cdot 398 = 19900$ .

Answer: a) $u_{n} = 4 n - 3$ ; b) $u_{100} = 397$ ; c) $S_{100} = 19900$

Exercise 15 BasicIB · P1 TZ2 May 2012

The tenth term of an arithmetic sequence is 32 and the common difference is $- 6$ .

a) Find the first term of the sequence.

b) Find the 21st term of the sequence.

c) Find the sum of the first 30 terms of the sequence.

Show answer

Answer: a) $u_{1} = 86$ . b) $u_{21} = - 34$ . c) $S_{30} = - 30$ .

Show working and answer

a) $u_{10} = u_{1} + 9 d$ , so $32 = u_{1} + 9 (- 6)$ and hence $u_{1} = 32 + 54 = 86$ .

b) $u_{21} = u_{1} + 20 d = 86 + 20 (- 6) = 86 - 120 = - 34$ .

c) $S_{30} = \frac{30}{2} (2 \cdot 86 + 29 \cdot (- 6)) = 15 (172 - 174) = 15 \cdot (- 2) = - 30$ .

Answer: a) $u_{1} = 86$ . b) $u_{21} = - 34$ . c) $S_{30} = - 30$ .

Exercise 16 BasicIB · P1 TZ0 November 2012

The first term of an arithmetic sequence is 3 and the seventh term is 33.

Calculate

a) the common difference;

b) the 95th term of the sequence;

c) the sum of the first 250 terms.

Show answer

Answer: a) $d = 5$ ; b) $u_{95} = 473$ ; c) $S_{250} = 156375$ .

Show working and answer

a) $u_{7} = u_{1} + 6 d$ , that is, $33 = 3 + 6 d$ , so $d = \frac{30}{6} = 5$ .

b) $u_{95} = u_{1} + 94 d = 3 + 94 \cdot 5 = 3 + 470 = 473$ .

c) $S_{250} = \frac{250}{2} (2 \cdot 3 + 249 \cdot 5) = 125 (6 + 1245) = 125 \cdot 1251 = 156375$ .

Answer: a) $d = 5$ ; b) $u_{95} = 473$ ; c) $S_{250} = 156375$ .

Exercise 17 BasicIB · P1 TZ2 May 2009

Consider the following sequence: $57, 55, 53, \dots, 5, 3$

a) Find the number of terms in the sequence.

b) Find the sum of the sequence.

Show answer

Answer: a) $28$ terms; b) $840$ .

Show working and answer

a) $d = 55 - 57 = - 2$ . The last term is $3$ :

$57 + (n - 1) (- 2) = 3$
$- 2 (n - 1) = - 54$
$n - 1 = 27 \Rightarrow n = 28$ .

b) $S_{28} = \frac{28}{2} (57 + 3) = 14 \cdot 60 = 840$ .

Answer: a) $28$ terms; b) $840$ .

Exercise 18 BasicIB · P1 TZ0 November 2016

A comet orbits the Sun and can be seen from Earth every 37 years. The first time the comet was seen from Earth was in the year 1064.

a) Find the year in which the comet was seen from Earth for the fifth time.

b) Determine how many times the comet has been seen from Earth up to the year 2014.

Show answer

Answer: a) year 1212. b) 26 times.

Show working and answer

The years of appearance form an arithmetic sequence with $u_{1} = 1064$ and $d = 37$ .

$u_{n} = u_{1} + (n - 1) d$ .
Fifth time: $u_{5} = 1064 + 4 \cdot 37 = 1064 + 148 = 1212$ .

We look for the largest $n$ with $u_{n} \leq 2014$ : $1064 + 37 (n - 1) \leq 2014$ .
$37 (n - 1) \leq 950 \Rightarrow n - 1 \leq 25.67 \Rightarrow n \leq 26.67$ .
As $n$ is an integer, $n = 26$ (in the year $1064 + 25 \cdot 37 = 1989$ was the last appearance before 2014).

Answer: a) year 1212. b) 26 times.

Exercise 19 MediumIB · P2 TZ0 November 2006

The first three terms of an arithmetic sequence are $2 k + 3, 5 k - 2$ , and $10 k - 15$ .

a) Show that $k = 4$ .

b) Find the values of the first three terms of the sequence.

c) Write down the value of the common difference.

d) Calculate the twentieth term of the sequence.

e) Find the sum of the first 15 terms of the sequence.

Show answer

Answer: b) $11, 18, 25$ ; c) $d = 7$ ; d) $u_{20} = 144$ ; e) $S_{15} = 900$

Show working and answer

a) In an arithmetic sequence the difference between consecutive terms is constant: $u_{2} - u_{1} = u_{3} - u_{2}$ .

$(5 k - 2) - (2 k + 3) = (10 k - 15) - (5 k - 2)$ .
We simplify each side: $3 k - 5 = 5 k - 13$ .
Collect terms: $8 = 2 k$ , so $k = 4$ .

as required.

b) With $k = 4$ :

$u_{1} = 2 \cdot 4 + 3 = 11$
$u_{2} = 5 \cdot 4 - 2 = 18$
$u_{3} = 10 \cdot 4 - 15 = 25$

c) $d = 18 - 11 = 7$ .

d) $u_{n} = u_{1} + (n - 1) d$ , so $u_{20} = 11 + 19 \cdot 7 = 11 + 133 = 144$ .

e) $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ :

$S_{15} = \frac{15}{2} (2 \cdot 11 + 14 \cdot 7) = \frac{15}{2} (22 + 98) = \frac{15}{2} \cdot 120 = 900.$

Answer: b) $11, 18, 25$ ; c) $d = 7$ ; d) $u_{20} = 144$ ; e) $S_{15} = 900$

Exercise 20 MediumIB · P2 TZ0 May 2004

The $n$ th term of an arithmetic sequence is given by $u_{n} = 63 - 4 n$ .

a) Calculate the values of the first two terms of this sequence.

b) Which term of the sequence is $- 13$ ?

c) The sum of two consecutive terms of the sequence, $u_{k}$ and $u_{k + 1}$ , is 34. Find $k$ .

Show answer

Answer: a) $u_{1} = 59, u_{2} = 55$ ; b) $n = 19$ ; c) $k = 11$

Show working and answer

$u_{1} = 63 - 4 \cdot 1 = 59$
$u_{2} = 63 - 4 \cdot 2 = 55$

b) We set $u_{n} = - 13$ : $63 - 4 n = - 13$ , so $4 n = 76$ and $n = 19$ .

$u_{k} = 63 - 4 k$ and $u_{k + 1} = 63 - 4 (k + 1) = 59 - 4 k$ .
Their sum: $(63 - 4 k) + (59 - 4 k) = 122 - 8 k$ .
We set $122 - 8 k = 34$ : $8 k = 88$ , so $k = 11$ .

Answer: a) $u_{1} = 59, u_{2} = 55$ ; b) $n = 19$ ; c) $k = 11$

Exercise 21 MediumIB · P1 TZ0 November 2004

The sixth term of an arithmetic sequence is 24. The common difference is 8.

a) Calculate the first term of the sequence.

The sum of the first $n$ terms is 600.

b) Calculate the value of $n$ .

Show answer

Answer: a) $u_{1} = - 16$ ; b) $n = 15$

Show working and answer

a) $u_{6} = u_{1} + 5 d$ , so $24 = u_{1} + 5 \cdot 8$ and $u_{1} = 24 - 40 = - 16$ .

$S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d) = \frac{n}{2} (- 32 + 8 (n - 1)) = \frac{n}{2} (8 n - 40) = 4 n^{2} - 20 n$ .
We set $S_{n} = 600$ : $4 n^{2} - 20 n - 600 = 0$ , that is, $n^{2} - 5 n - 150 = 0$ .
Factorise: $(n - 15) (n + 10) = 0$ , so $n = 15$ or $n = - 10$ .
As $n$ is a natural number, $n = 15$ .

Answer: a) $u_{1} = - 16$ ; b) $n = 15$

Exercise 22 MediumIB · P1 TZ1 May 2013

In an arithmetic sequence, the fifth term, $u_{5}$ , is greater than the first term, $u_{1}$ .

The difference between these terms is 36.

a) Find the common difference, $d$ .

The tenth term of the sequence is double the seventh term.

b) (i) Write down an equation in $u_{1}$ and $d$ to show this information.

(ii) Calculate $u_{1}$ .

Show answer

Answer: a) $d = 9$ ; b) (i) $u_{1} + 9 d = 2 (u_{1} + 6 d)$ ; (ii) $u_{1} = - 27$

Show working and answer

a) $u_{5} = u_{1} + 4 d$ , so $u_{5} - u_{1} = 4 d = 36$ and $d = 9$ .

b) (i) The tenth term is $u_{10} = u_{1} + 9 d$ and the seventh is $u_{7} = u_{1} + 6 d$ . The tenth being double the seventh is written as:

$u_{1} + 9 d = 2 (u_{1} + 6 d) .$

(ii) 1. We substitute $d = 9$ : $u_{1} + 81 = 2 (u_{1} + 54) = 2 u_{1} + 108$ . 2. Collect terms: $81 - 108 = 2 u_{1} - u_{1}$ , so $u_{1} = - 27$ .

Check: $u_{7} = - 27 + 54 = 27$ and $u_{10} = - 27 + 81 = 54 = 2 \cdot 27$ .

Answer: a) $d = 9$ ; b) (i) $u_{1} + 9 d = 2 (u_{1} + 6 d)$ ; (ii) $u_{1} = - 27$

Exercise 23 MediumIB · P1 TZ2 May 2010

A choir is getting ready to give a concert. Its members stand in rows, arranged in an arithmetic sequence. So, in the fourth row there are 20 singers, and in the eighth row there are 32 singers.

a) Find the common difference of this arithmetic sequence.

The choir has 10 rows. In the first row there are 11 singers.

b) Find the total number of singers in the choir.

Show answer

Answer: a) $d = 3$ . b) $245$ singers.

Show working and answer

From the eighth term to the fourth term there are $4$ steps of the common difference: $u_{8} = u_{4} + 4 d$ .
$32 = 20 + 4 d$ , so $4 d = 12$ and $d = 3$ .

With $u_{1} = 11$ , $d = 3$ and $n = 10$ : $S_{10} = \frac{10}{2} (2 \cdot 11 + 9 \cdot 3) = 5 (22 + 27) = 5 \cdot 49 = 245$ .
Check with the data: $u_{4} = 11 + 9 = 20$ and $u_{8} = 11 + 21 = 32$ , both correct.

Answer: a) $d = 3$ . b) $245$ singers.

Exercise 24 MediumIB · P1 TZ0 November 2010

The first term of an arithmetic sequence is 3 and the sum of the first two terms is equal to 11.

a) Write down the second term of this sequence.

b) Write down the common difference of this sequence.

c) Write down the fourth term of this sequence.

d) The $n$ th term is the first term of this sequence that is greater than 1000.

Find the value of $n$ .

Show answer

Answer: a) $u_{2} = 8$ ; b) $d = 5$ ; c) $u_{4} = 18$ ; d) $n = 201$ .

Show working and answer

a) The sum of the first two terms is $u_{1} + u_{2} = 11$ and $u_{1} = 3$ , so $u_{2} = 11 - 3 = 8$ .

b) The common difference is $d = u_{2} - u_{1} = 8 - 3 = 5$ .

c) $u_{4} = u_{1} + 3 d = 3 + 3 \cdot 5 = 18$ .

d) The general term is $u_{n} = 3 + 5 (n - 1)$ . We require $u_{n} > 1000$ :

$3 + 5 (n - 1) > 1000$
$5 (n - 1) > 997$
$n - 1 > 199.4$ , that is, $n > 200.4$ .

As $n$ is an integer, the first possible value is $n = 201$ . Check: $u_{201} = 3 + 5 \cdot 200 = 1003 > 1000$ and $u_{200} = 998 < 1000$ .

Answer: a) $u_{2} = 8$ ; b) $d = 5$ ; c) $u_{4} = 18$ ; d) $n = 201$ .

Exercise 25 MediumIB · P1 TZ0 November 2011

Consider the arithmetic sequence $326, 321, 316, 311, \dots, 191$ .

a) Find the value of the common difference of this sequence.

b) Calculate the sum of the first 10 terms of this sequence.

c) Find the number of terms in this sequence.

Show answer

Answer: a) $d = - 5$ ; b) $S_{10} = 3035$ ; c) $28$ terms.

Show working and answer

a) $d = 321 - 326 = - 5$ .

b) $S_{10} = \frac{10}{2} (2 \cdot 326 + 9 \cdot (- 5)) = 5 (652 - 45) = 5 \cdot 607 = 3035$ .

c) The last term is $191$ , so $u_{n} = 326 + (n - 1) (- 5) = 191$ :

$- 5 (n - 1) = 191 - 326 = - 135$
$n - 1 = 27$
$n = 28$ .

Answer: a) $d = - 5$ ; b) $S_{10} = 3035$ ; c) $28$ terms.

Exercise 26 MediumIB · P1 TZ1 May 2009

A tree starts to lose its leaves in October. The number of leaves the tree loses increases by the same amount each day.

October day	1	2	3	4	.....................
Number of leaves lost	24	40	56	72	.....................

a) Calculate the number of leaves the tree loses on 21 October.

b) Find the total number of leaves the tree loses over the 31 days of October.

Show answer

Answer: a) $344$ leaves; b) $8184$ leaves.

Show working and answer

The quantities $24, 40, 56, 72, \dots$ form an arithmetic sequence with $u_{1} = 24$ and $d = 40 - 24 = 16$ .

a) 21 October corresponds to $n = 21$ : $u_{21} = 24 + 20 \cdot 16 = 24 + 320 = 344$ leaves.

b) October has 31 days: $S_{31} = \frac{31}{2} (2 \cdot 24 + 30 \cdot 16) = \frac{31}{2} \cdot 528 = 31 \cdot 264 = 8184$ leaves.

Answer: a) $344$ leaves; b) $8184$ leaves.

Exercise 27 MediumIB · P1 TZ0 November 2009

Given that the first term of an arithmetic sequence is 7 and the sixth term is 22, find

a) the common difference;

b) the twelfth term (the number 12);

c) the sum of the first 100 terms.

Show answer

Answer: $d = 3$ ; $u_{12} = 40$ ; $S_{100} = 15550$ .

Show working and answer

a) In an arithmetic sequence $u_{n} = u_{1} + (n - 1) d$ . With $u_{6} = 22$ and $u_{1} = 7$ :

$7 + 5 d = 22$ .
$5 d = 15$ , so $d = 3$ .

Answer a): $d = 3$ .

b) 1. $u_{12} = u_{1} + 11 d = 7 + 11 \cdot 3$ . 2. $u_{12} = 7 + 33 = 40$ .

Answer b): $u_{12} = 40$ .

c) 1. Sum formula: $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ . 2. $S_{100} = 50 (14 + 99 \cdot 3) = 50 \cdot 311$ . 3. $S_{100} = 15550$ .

Answer c): $S_{100} = 15550$ .

Answer: $d = 3$ ; $u_{12} = 40$ ; $S_{100} = 15550$ .

Exercise 28 MediumIB · P1 TZ0 November 2008

Given the arithmetic sequence: $u_{1} = 124, u_{2} = 117, u_{3} = 110, u_{4} = 103, \dots$

a) Write down the value of the common difference of this sequence.

b) Calculate the sum of the first 50 terms of the sequence.

$u_{k}$ is the first term of the sequence that is negative.

c) Find the value of $k$ .

Show answer

Answer: $d = - 7$ ; $S_{50} = - 2375$ ; $k = 19$ .

Show working and answer

a) $d = u_{2} - u_{1} = 117 - 124 = - 7$ .

Answer a): $d = - 7$ .

b) 1. $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ . 2. $S_{50} = 25 (248 + 49 \cdot (- 7)) = 25 (248 - 343)$ . 3. $S_{50} = 25 \cdot (- 95) = - 2375$ .

Answer b): $S_{50} = - 2375$ .

c) 1. $u_{n} = 124 - 7 (n - 1) = 131 - 7 n$ . 2. We look for the first $n$ with $u_{n} < 0$ : $131 - 7 n < 0 \Rightarrow n > \frac{131}{7} \approx 18.71$ . 3. The first integer is $n = 19$ . Check: $u_{18} = 5 > 0$ and $u_{19} = - 2 < 0$ .

Answer c): $k = 19$ .

Answer: $d = - 7$ ; $S_{50} = - 2375$ ; $k = 19$ .

Exercise 29 MediumIB · P2 TZ0 May 2007

The natural numbers $1, 2, 3, 4, 5 \dots$ form an arithmetic sequence.

a) State the values of $u_{1}$ and $d$ for this sequence.

b) Use an appropriate formula to prove that the sum of the natural numbers from 1 to $n$ is given by $\frac{1}{2} n (n + 1)$ .

c) Calculate the sum of the natural numbers from 1 to 200.

Show answer

Answer: $u_{1} = 1$ , $d = 1$ ; formula proved; $S_{200} = 20100$ .

Show working and answer

a) The sequence $1, 2, 3, 4, \dots$ starts at 1, and each term adds 1 to the previous one: $u_{1} = 1$ , $d = 1$ .

Answer a): $u_{1} = 1$ ; $d = 1$ .

b) 1. We use $S_{n} = \frac{n}{2} (2 u_{1} + (n - 1) d)$ . 2. Substituting $u_{1} = 1$ and $d = 1$ : $S_{n} = \frac{n}{2} (2 + (n - 1))$ . 3. Simplifying: $S_{n} = \frac{n}{2} (n + 1) = \frac{1}{2} n (n + 1)$ .

Proved.

c) 1. $S_{200} = \frac{1}{2} \cdot 200 \cdot 201$ . 2. $S_{200} = 100 \cdot 201 = 20100$ .

Answer c): $S_{200} = 20100$ .

Answer: $u_{1} = 1$ , $d = 1$ ; formula proved; $S_{200} = 20100$ .

Exercise 30 MediumIB · P1 TZ0 November 2007

The fifth term of an arithmetic sequence is 20 and the twelfth term is 41.

a) (i) Find the common difference.

(ii) Find the first term of the sequence.

b) Calculate the 84th term.

c) Calculate the sum of the first 200 terms.

Show answer

Answer: $d = 3$ ; $u_{1} = 8$ ; $u_{84} = 257$ ; $S_{200} = 61300$ .

Show working and answer

a) We set up $u_{5} = u_{1} + 4 d = 20$ and $u_{12} = u_{1} + 11 d = 41$ .

Subtract the first equation from the second: $7 d = 21$ .
$d = 3$ .
Substituting into the first: $u_{1} + 12 = 20$ , so $u_{1} = 8$ .

Answer a): (i) $d = 3$ ; (ii) $u_{1} = 8$ .

b) 1. $u_{84} = u_{1} + 83 d = 8 + 83 \cdot 3$ . 2. $u_{84} = 8 + 249 = 257$ .

Answer b): $u_{84} = 257$ .

c) 1. $S_{200} = 100 (2 \cdot 8 + 199 \cdot 3)$ . 2. $S_{200} = 100 (16 + 597) = 100 \cdot 613 = 61300$ .

Answer c): $S_{200} = 61300$ .

Answer: $d = 3$ ; $u_{1} = 8$ ; $u_{84} = 257$ ; $S_{200} = 61300$ .

Exercise 31 MediumIB · P1 TZ2 May 2017

Tomás is playing with some sticks and uses them to form the first three diagrams of a pattern. These diagrams are shown below.

Three stick diagrams: Diagram 1, a square of 4 sticks; Diagram 2, two joined squares of 7 sticks; Diagram 3, three joined squares of 10 sticks

Tomás continues forming diagrams following this same pattern.

a) To form diagram $n$ he uses 52 sticks. Find the value of $n$ .

Tomás forms 24 diagrams in total.

b) Find the total number of sticks that Tomás has used to form these 24 diagrams.

Show answer

Answer: a) $n = 17$ . b) 924 sticks.

Show working and answer

The numbers of sticks $4, 7, 10, \dots$ form an arithmetic sequence with $u_{1} = 4$ and $d = 3$ .

$u_{n} = 4 + (n - 1) \cdot 3 = 3 n + 1$ .
$3 n + 1 = 52 \Rightarrow 3 n = 51 \Rightarrow n = 17$ .

$u_{24} = 3 \cdot 24 + 1 = 73$ .
$S_{24} = \frac{24}{2} (4 + 73) = 12 \cdot 77 = 924$ sticks.

Answer: a) $n = 17$ . b) 924 sticks.

Exercise 32 MediumIB · P1 TZ1 May 2016

One of the venues for the 2016 Olympic Games is an amphitheatre. The number of seats in the first row of the amphitheatre, $u_{1}$ , is 240. The number of seats in each subsequent row forms an arithmetic sequence. The number of seats in the sixth row, $u_{6}$ , is 270.

a) Calculate the value of the common difference, $d$ .

There are 20 rows in the amphitheatre.

b) Find the total number of seats in the amphitheatre.

Anisha visits the amphitheatre. She estimates that the amphitheatre has 6,500 seats.

c) Calculate the percentage error in Anisha's estimate.

Show answer

Answer: a) $d = 6$ . b) 5,940 seats. c) $\approx 9.43 %$ .

Show working and answer

$u_{6} = u_{1} + 5 d \Rightarrow 270 = 240 + 5 d$ .
$5 d = 30 \Rightarrow d = 6$ .

$S_{20} = \frac{20}{2} [2 \cdot 240 + 19 \cdot 6] = 10 \cdot (480 + 114) = 10 \cdot 594 = 5940$ seats.

Percentage error $= \frac{∣ estimated value - exact value ∣}{∣ exact value ∣} \cdot 100$ .
$\frac{∣6500 - 5940∣}{5940} \cdot 100 = \frac{560}{5940} \cdot 100 \approx 9.43 %$ .

Answer: a) $d = 6$ . b) 5,940 seats. c) $\approx 9.43 %$ .

Exercise 33 MediumIB · P1 TZ1 May 2017

The company Snakezen's Ladders manufactures ladders of different lengths. All the ladders the company makes have the same design, such that:

the first rung is 30 cm from the base of the ladder, the second rung is 57 cm from the base of the ladder, the distance between the first and second rungs is equal to the distance between all adjacent rungs on the ladder.

The ladder in the diagram was made by this company and has eleven equally spaced rungs.

Vertical ladder with two stiles and eleven equally spaced rungs; a dimension labelled 30 cm marks the distance from the base to the first rung and another labelled 57 cm marks the distance from the base to the second rung

a) Find the distance from the base of this ladder to the top rung.

The company also manufactures a ladder that is 1,050 cm long.

b) Find the maximum number of rungs on this 1,050 cm long ladder.

Show answer

Answer: the maximum number of rungs is $38$ .

Show working and answer

The heights of the rungs form an arithmetic sequence: $u_{1} = 30$ and $u_{2} = 57$ .
The difference is $d = 57 - 30 = 27$ cm.
The top rung is the eleventh: $u_{11} = u_{1} + (11 - 1) d = 30 + 10 \cdot 27 = 300$ .

Answer a): the distance is $300$ cm.

Rung $n$ is $u_{n} = 30 + 27 (n - 1)$ cm from the base, and we need $u_{n} \leq 1050$ .
$30 + 27 (n - 1) \leq 1050 \Rightarrow 27 (n - 1) \leq 1020 \Rightarrow n - 1 \leq 37.78 \dots$
Since $n$ is an integer, $n - 1 = 37$ , so $n = 38$ . Check: $u_{38} = 30 + 27 \cdot 37 = 1029 \leq 1050$ , and $u_{39} = 1056 > 1050$ .

Answer: the maximum number of rungs is $38$ .

Exercise 34 MediumIB · P1 TZ1 May 2018

Sergei is training to be a weightlifter. Each day he trains at the local gym, lifting a bar with heavy weights attached. He performs successive lifts. After each lift, he adds an amount of weight to the bar to increase the weight to be lifted. The weights of Sergei's lifts form an arithmetic sequence. Sergei's friend, Yuri, records the weight of each lift. Unfortunately, last Monday, Yuri lost all but two of the records of Sergei's lifts. That day, Sergei lifted 21 kg on the third lift and 46 kg on the eighth lift.

a) For that day,

(i) find how much weight was added after each lift;

(ii) find the weight of Sergei's first lift.

That day, Sergei performed 12 successive lifts.

b) Find the total combined weight of these lifts.

View solution

The weights form an arithmetic sequence with $u_{3} = 21$ and $u_{8} = 46$ .
Between the third and eighth lifts there are $5$ increments: $u_{8} - u_{3} = 5 d$ .

(i) $5 d = 46 - 21 = 25 \Rightarrow d = 5$ .

(ii) $u_{1} = u_{3} - 2 d = 21 - 10 = 11$ .

Answer a): (i) $5$ kg was added after each lift; (ii) the first lift was $11$ kg.

$u_{12} = 11 + 11 \cdot 5 = 66$ .
$S_{12} = \frac{12}{2} (u_{1} + u_{12}) = 6 (11 + 66) = 462$ .

Answer b): the total combined weight is $462$ kg.

Exercise 35 MediumIB · P1 TZ0 November 2000

Mr Jones decides to increase the amount of money he spends on food by $d$ GBP each year. In the first year he spends $a$ GBP. In the 8th year he spends twice as much as in the 4th year. In the 20th year he spends 4,000 GBP.

Find the value of $d$ .

Show answer

Answer: $d = 200$ GBP.

Show working and answer

The expenses form an arithmetic sequence with first term $a$ and common difference $d$ : $u_{n} = a + (n - 1) d$ .

Year 4: $a + 3 d$ . Year 8: $a + 7 d$ . As year 8 is twice year 4: $a + 7 d = 2 (a + 3 d)$ .
Simplifying: $a + 7 d = 2 a + 6 d \Rightarrow a = d$ .
Year 20: $a + 19 d = 4000$ . Substituting $a = d$ : $20 d = 4000$ .
Therefore $d = 200$ .

Answer: $d = 200$ GBP.

Exercise 36 HardIB · P2 TZ2 May 2015

The sum of the first $n$ terms of an arithmetic sequence is given by $S_{n} = 6 n + n^{2}$ .

a) Write down the value of

(i) $S_{1}$ ;

(ii) $S_{2}$ .

The $n$ th term of the arithmetic sequence is denoted by $u_{n}$ .

b) Show that $u_{2} = 9$ .

c) Find the common difference of the sequence.

d) Calculate $u_{10}$ .

e) Find the smallest value of $n$ for which $u_{n}$ is greater than 1,000.

f) There is a value of $n$ for which $u_{1} + u_{2} + \dots + u_{n} = 1512$

Calculate the value of $n$ .

Show answer

Answer: a) $S_{1} = 7$ , $S_{2} = 16$ . b) $u_{2} = 9$ (shown). c) $d = 2$ . d) $u_{10} = 25$ . e) $n = 498$ . f) $n = 36$ .

Show working and answer

(i) $S_{1} = 6 \cdot 1 + 1^{2} = 7$ .
(ii) $S_{2} = 6 \cdot 2 + 2^{2} = 16$ .

b) $u_{1} = S_{1} = 7$ . Since $S_{2} = u_{1} + u_{2}$ : $u_{2} = S_{2} - S_{1} = 16 - 7 = 9$ as required.

c) $d = u_{2} - u_{1} = 9 - 7 = 2$ .

d) $u_{10} = u_{1} + 9 d = 7 + 9 \cdot 2 = 25$ .

$u_{n} = 7 + (n - 1) \cdot 2 = 2 n + 5$ .
We require $2 n + 5 > 1000$ , that is $n > 497.5$ .
The smallest integer satisfying this is $n = 498$ (check: $u_{498} = 1001 > 1000$ and $u_{497} = 999$ ).

We need $S_{n} = 1512$ , that is $6 n + n^{2} = 1512$ , or $n^{2} + 6 n - 1512 = 0$ .
Discriminant: $36 + 4 \cdot 1512 = 6084 = 7 8^{2}$ .
$n = \frac{- 6 \pm 78}{2}$ , so $n = 36$ or $n = - 42$ . As $n$ is positive, $n = 36$ .

Answer: a) $S_{1} = 7$ , $S_{2} = 16$ . b) $u_{2} = 9$ (shown). c) $d = 2$ . d) $u_{10} = 25$ . e) $n = 498$ . f) $n = 36$ .

Exercise 37 HardIB · P2 TZ0 November 2017

Rosa joins a club where people train to run a marathon. During the first training session Rosa runs a distance of 3,000 metres. In each training session Rosa runs 400 metres more than in the previous one.

a) Write down the distance Rosa runs

(i) in the third training session;

(ii) in the $n$ th training session.

A marathon is 42.195 kilometres. In the $k$ th training session Rosa will run a distance greater than a marathon for the first time.

b) Find the value of $k$ .

c) Calculate the total distance, in kilometres, that Rosa will have run in the first 50 training sessions.

Carlos joins the club to lose weight. In the first month he runs 7,500 metres. The distance Carlos runs increases by 20% each month.

d) Find the distance Carlos runs during the fifth month of training.

e) Calculate the total distance Carlos runs during the first year.

Show answer

Answer: a) (i) 3,800 m; (ii) $3000 + 400 (n - 1)$ m. b) $k = 99$ . c) 640 km. d) 15,552 m ( $\approx 15600$ m). e) $\approx 296854$ m ( $\approx 297000$ m).

Show working and answer

a) Distances in metres: arithmetic sequence with $u_{1} = 3000$ and $d = 400$ .

(i) $u_{3} = 3000 + 2 \cdot 400 = 3800$ m.
(ii) $u_{n} = 3000 + (n - 1) \cdot 400 = 2600 + 400 n$ m.

A marathon is $42195$ m. We require $2600 + 400 k > 42195$ .
$400 k > 39595 \Rightarrow k > 98.99$ .
The smallest integer is $k = 99$ . Check: $u_{99} = 42200 > 42195$ and $u_{98} = 41800 < 42195$ .

$S_{50} = \frac{50}{2} [2 \cdot 3000 + 49 \cdot 400] = 25 \cdot 25600 = 640000$ m.
In kilometres: $640$ km.

d) Carlos's distances: geometric sequence with $v_{1} = 7500$ and $r = 1.2$ .

$v_{5} = 7500 \cdot 1. 2^{4} = 7500 \cdot 2.0736 = 15552$ m ( $\approx 15600$ m).

First year $= 12$ months: $S_{12} = \frac{7500 ( 1. 2 ^{12} - 1 )}{0.2}$ .
$S_{12} \approx 296854$ m ( $\approx 297000$ m).

Answer: a) (i) 3,800 m; (ii) $3000 + 400 (n - 1)$ m. b) $k = 99$ . c) 640 km. d) 15,552 m ( $\approx 15600$ m). e) $\approx 296854$ m ( $\approx 297000$ m).

Exercise 38 HardIB · P2 TZ0 November 2014

In a game, $n$ small pumpkins are placed 1 metre apart in a straight line. The players start 3 metres before the first pumpkin.

Horizontal line starting from a black dot labelled Start, on the left, passing through a row of equally spaced pumpkins, followed by suspension points on the right

Each player collects a single pumpkin and returns it to the start. The nearest pumpkin is collected first. The player then collects the next nearest pumpkin, and the game continues in this way until the signal to end is given. Sirma runs to fetch each pumpkin and returns it to the start.

a) Write down the distance, $a_{1}$ , in metres that she has to run to collect the first pumpkin.

The distances she runs to collect each pumpkin form a sequence $a_{1}, a_{2}, a_{3}, \dots$

b) (i) Calculate $a_{2}$ .

(ii) Calculate $a_{3}$ .

c) Write down the common difference, $d$ , of the sequence.

The final pumpkin that Sirma collected was 24 metres from the start.

d) (i) Find the total number of pumpkins that Sirma collected.

(ii) Find the total distance that Sirma ran to collect these pumpkins.

Peter also plays the game. When the signal to end the game is given, he has run 940 metres.

e) Calculate the total number of pumpkins that Peter collected.

f) Calculate the distance between Peter and the starting point of the game when the signal is given.

Show answer

Answer: a) $a_{1} = 6$ m. b) (i) $a_{2} = 8$ m; (ii) $a_{3} = 10$ m. c) $d = 2$ m. d) (i) 22 pumpkins; (ii) 594 m. e) 28 pumpkins. f) 16 m.

Show working and answer

a) The first pumpkin is 3 m from the start: going and returning is $2 \cdot 3 = 6$ m. So $a_{1} = 6$ m.

The second pumpkin is $4$ m away: $a_{2} = 2 \cdot 4 = 8$ m.
The third is $5$ m away: $a_{3} = 2 \cdot 5 = 10$ m.

c) $d = a_{2} - a_{1} = 8 - 6 = 2$ m (each pumpkin adds 1 m going and 1 m returning).

(i) It is an arithmetic sequence: $a_{n} = 6 + (n - 1) \cdot 2$ . The last pumpkin is 24 m away, so its round trip is $48$ m: $6 + 2 (n - 1) = 48 \Rightarrow n - 1 = 21 \Rightarrow n = 22$ pumpkins.
(ii) $S_{22} = \frac{22}{2} (6 + 48) = 11 \cdot 54 = 594$ m.

The total distance after $n$ complete pumpkins is $S_{n} = \frac{n}{2} [12 + 2 (n - 1)] = n (n + 5)$ .
We solve $n (n + 5) \leq 940$ (using the GDC or with $n^{2} + 5 n - 940 = 0$ , which gives $n \approx 28.3$ ).
$S_{28} = 28 \cdot 33 = 924 \leq 940$ and $S_{29} = 29 \cdot 34 = 986 > 940$ . Therefore Peter collected $28$ complete pumpkins.

After 28 pumpkins he has run 924 m; he has $940 - 924 = 16$ m left on the next trip (pumpkin 29 is 31 m from the start, so he is still on the outward leg).
Therefore he is $16$ m from the starting point.

Answer: a) $a_{1} = 6$ m. b) (i) $a_{2} = 8$ m; (ii) $a_{3} = 10$ m. c) $d = 2$ m. d) (i) 22 pumpkins; (ii) 594 m. e) 28 pumpkins. f) 16 m.

Exercise 39 HardIB · P2 TZ0 November 2019

Maegan designs a decorative stained-glass window for the facade of a new Fine Arts Centre. The window is made of small triangular panels. The first three levels of the window are shown in the following figure.

Stained-glass window in the shape of an inverted trapezium with triangular panels in three levels: the first level (bottom) has 5 panels, the second level 7 panels and the third level (top) 9 panels; two dashed lines extend upwards from the top ends

The 1st level — the one at the bottom of the window — consists of 5 triangular panels. The 2nd level has 7 triangular panels and the 3rd level has 9 triangular panels. Each additional level has 2 more triangular panels than the level below it.

a) Find the number of triangular panels in the 12th level.

b) Show that the total number of triangular panels, $S_{n}$ , in the first $n$ levels is given by: $S_{n} = n^{2} + 4 n$ .

c) Hence, find the total number of panels in a window with 18 levels.

Maegan has 1,000 triangular panels to build the window and does not want to leave any level incomplete.

d) Find the maximum number of complete levels that Maegan can make.

The area of each triangular panel is equal to 1.84 m $^{2}$ .

e) Find the total area of the window if it is built with that maximum number of complete levels. Give your answer correct to the nearest m $^{2}$ .

Show answer

Answer: a) 27 panels. b) shown. c) 396 panels. d) 29 levels. e) $\approx 1761$ m $^{2}$ .

Show working and answer

Panels per level: $5, 7, 9, \dots$ , an arithmetic sequence with $u_{1} = 5$ and $d = 2$ .

$u_{12} = 5 + 11 \cdot 2 = 27$ panels.

$S_{n} = \frac{n}{2} [2 u_{1} + (n - 1) d] = \frac{n}{2} [10 + 2 (n - 1)]$ .
Simplifying: $\frac{n}{2} (2 n + 8) = n (n + 4) = n^{2} + 4 n$ .

This completes the proof.

$S_{18} = 1 8^{2} + 4 \cdot 18 = 324 + 72 = 396$ panels.

We look for the largest $n$ with $n^{2} + 4 n \leq 1000$ .
$n = 29$ : $841 + 116 = 957 \leq 1000$ . $n = 30$ : $900 + 120 = 1020 > 1000$ .
Maximum: $29$ levels.

Panels used: $S_{29} = 957$ .
Area: $957 \cdot 1.84 = 1760.88 \approx 1761$ m $^{2}$ .

Answer: a) 27 panels. b) shown. c) 396 panels. d) 29 levels. e) $\approx 1761$ m $^{2}$ .

Geometric sequences: term and sum

Exercise 40 BasicIB · P1 TZ0 November 2006

Consider the geometric sequence $8, a, 2, \dots$ for which the common ratio is $\frac{1}{2}$ .

a) Find the value of $a$ .

b) Find the value of the eighth term.

c) Find the sum of the first twelve terms.

Show answer

Answer: a) $a = 4$ ; b) $u_{8} = \frac{1}{16} = 0.0625$ ; c) $S_{12} \approx 16.0$

Show working and answer

a) In a geometric sequence, each term is the previous term multiplied by the common ratio: $a = 8 \cdot \frac{1}{2} = 4$ . (Check: $4 \cdot \frac{1}{2} = 2$ .)

b) We use $u_{n} = u_{1} r^{n - 1}$ with $u_{1} = 8$ , $r = \frac{1}{2}$ , $n = 8$ :

$u_{8} = 8 (\frac{1}{2})^{7} = \frac{8}{128} = \frac{1}{16} = 0.0625.$

c) Sum of the first $n$ terms: $S_{n} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ .

$S_{12} = \frac{8 ( 1 - ( \frac{1}{2} ) ^{12} )}{1 - \frac{1}{2}} = 16 (1 - \frac{1}{4096}) \approx 15.996 \approx 16.0.$

Answer: a) $a = 4$ ; b) $u_{8} = \frac{1}{16} = 0.0625$ ; c) $S_{12} \approx 16.0$

Exercise 41 MediumIB · P1 TZ0 November 2019

A geometric sequence has first term $\frac{8}{3}$ and fourth term 9.

a) Find the common ratio of the sequence.

b) Write down the second term of this sequence.

The sum of the first $k$ terms is greater than 2500.

c) Find the smallest possible value of $k$ .

Show answer

Answer: a) $r = \frac{3}{2}$ ; b) $u_{2} = 4$ ; c) $k = 16$

Show working and answer

a) In a geometric sequence $u_{n} = u_{1} r^{n - 1}$ , so $u_{4} = u_{1} r^{3}$ .

We substitute the given values: $9 = \frac{8}{3} r^{3}$ .
We solve for $r^{3}$ : $r^{3} = 9 \cdot \frac{3}{8} = \frac{27}{8}$ .
Take the cube root: $r = \frac{3}{2}$ .

Answer: $r = \frac{3}{2}$

b) The second term is $u_{2} = u_{1} r = \frac{8}{3} \cdot \frac{3}{2} = 4$ .

Answer: $u_{2} = 4$

c) We use the sum of the first $k$ terms, $S_{k} = \frac{u _{1} ( r ^{k} - 1 )}{r - 1}$ .

$S_{k} = \frac{\frac{8}{3} ( 1. 5 ^{k} - 1 )}{0.5} = \frac{16}{3} (1. 5^{k} - 1)$ .
We require $S_{k} > 2500$ : $\frac{16}{3} (1. 5^{k} - 1) > 2500$ , that is, $1. 5^{k} - 1 > 468.75$ and hence $1. 5^{k} > 469.75$ .
Take logarithms: $k > \frac{ln 469.75}{ln 1.5} \approx 15.2$ .
As $k$ is an integer, the smallest value is $k = 16$ .

Check: $S_{15} = \frac{16}{3} (1. 5^{15} - 1) \approx 2330$ , which does not reach 2500, and $S_{16} = \frac{16}{3} (1. 5^{16} - 1) \approx 3498$ , which does exceed it.

Answer: a) $r = \frac{3}{2}$ ; b) $u_{2} = 4$ ; c) $k = 16$

Exercise 42 MediumIB · P1 TZ1 May 2017

The first three terms of a geometric sequence are $u_{1} = 486, u_{2} = 162, u_{3} = 54$ .

a) Find the value of $r$ , the common ratio of the sequence.

b) Find the value of $n$ for which $u_{n} = 2$ .

c) Find the sum of the first 30 terms of the sequence.

Show answer

Answer: a) $r = \frac{1}{3}$ ; b) $n = 6$ ; c) $S_{30} \approx 729$

Show working and answer

a) $r = \frac{u _{2}}{u _{1}} = \frac{162}{486} = \frac{1}{3}$ . (Check: $162 \cdot \frac{1}{3} = 54$ .)

$u_{n} = 486 (\frac{1}{3})^{n - 1} = 2$ .
$(\frac{1}{3})^{n - 1} = \frac{2}{486} = \frac{1}{243} = \frac{1}{3 ^{5}}$ .
Equate the exponents: $n - 1 = 5$ , so $n = 6$ .

c) $S_{n} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ :

$S_{30} = \frac{486 ( 1 - ( \frac{1}{3} ) ^{30} )}{\frac{2}{3}} = 729 (1 - 3^{- 30}) \approx 729.$

(The term $3^{- 30}$ is of the order of $1 0^{- 15}$ , and does not affect the result.)

Answer: a) $r = \frac{1}{3}$ ; b) $n = 6$ ; c) $S_{30} \approx 729$

Exercise 43 MediumIB · P1 TZ0 November 2015

Consider the geometric sequence $2048, 1536, 1152, 864 \dots$

a) Find the common ratio, $r$ .

b) Write down the next term of the sequence, $u_{5}$ .

c) Find the largest term of the sequence that is not an integer.

d) Find the sum of the first 20 terms.

Show answer

Answer: a) $r = 0.75$ . b) $u_{5} = 648$ . c) $u_{7} = 364.5$ . d) $S_{20} \approx 8166$ (approximately $8170$ to 3 significant figures).

Show working and answer

a) $r = \frac{1536}{2048} = \frac{3}{4} = 0.75$ . Check: $\frac{1152}{1536} = 0.75$ and $\frac{864}{1152} = 0.75$ .

b) $u_{5} = u_{4} \cdot r = 864 \cdot 0.75 = 648$ .

The general term is $u_{n} = 2048 \cdot (\frac{3}{4})^{n - 1} = 2^{11} \cdot \frac{3 ^{n - 1}}{2 ^{2 (n - 1)}} = 3^{n - 1} \cdot 2^{13 - 2 n}$ .
While $13 - 2 n \geq 0$ , that is $n \leq 6$ , the term is an integer. So $u_{6} = 3^{5} \cdot 2 = 486$ .
For $n = 7$ : $u_{7} = 3^{6} \cdot 2^{- 1} = \frac{729}{2} = 364.5$ , which is not an integer.
For $n \geq 7$ the denominator is a power of 2 ( $2^{2 n - 13}$ ) and the numerator is a power of 3 (odd), so no later term is an integer.
The sequence is decreasing, so the largest non-integer term is the first one that stops being an integer: $u_{7} = 364.5$ .

d) Using $S_{n} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ : $S_{20} = \frac{2048 ( 1 - 0.7 5 ^{20} )}{1 - 0.75} = 8192 (1 - 0.7 5^{20}) \approx 8192 (1 - 0.003171) \approx 8166$

Answer: a) $r = 0.75$ . b) $u_{5} = 648$ . c) $u_{7} = 364.5$ . d) $S_{20} \approx 8166$ (approximately $8170$ to 3 significant figures).

Exercise 44 MediumIB · P2 TZ0 May 2004

A basketball is dropped vertically. On its first bounce it reaches a height of 2 m. The height of each subsequent bounce is 90% of the previous one.

a) What height will it reach on the 8th bounce?

b) What is the total vertical distance travelled by the ball between the first and the sixth time it touches the ground?

Show answer

Answer: a) $0.957$ m. b) $\approx 16.4$ m.

Show working and answer

The bounce heights form a geometric sequence with first term $2$ (first bounce) and common ratio $r = 0.9$ .

a) Height of the 8th bounce: $u_{8} = 2 \cdot 0. 9^{7} \approx 2 \cdot 0.4783 = 0.957$ m.

Between the first and the sixth time the ball touches the ground there are five complete bounces (bounces 1 to 5), each consisting of an upward and a downward movement.
Heights of these bounces: $2; 1.8; 1.62; 1.458; 1.3122$ . Each bounce covers twice its height.
Sum of the heights, with $S_{5} = \frac{2 ( 1 - 0. 9 ^{5} )}{1 - 0.9} = 20 (1 - 0.59049) = 8.1902$ m.
Total distance: $2 \cdot 8.1902 = 16.3804 \approx 16.4$ m.

Answer: a) $0.957$ m. b) $\approx 16.4$ m.

Exercise 45 MediumIB · P1 TZ2 May 2012

Javier starts training for a running race.

On the first day he runs 1.5 km. Every day he runs 10% more than the previous day.

a) Write down the distance he runs on the second day of training.

b) Calculate the total distance Javier runs in the first seven days of training.

Javier stops training on the day his total distance exceeds 100 km.

c) Calculate the number of days Javier has trained for the race.

Show answer

Answer: a) $1.65$ km. b) $\approx 14.2$ km. c) He trains for $22$ days.

Show working and answer

The daily distances form a geometric sequence with $u_{1} = 1.5$ and $r = 1.10$ (10% more each day).

a) $u_{2} = 1.5 \cdot 1.1 = 1.65$ km.

b) $S_{7} = \frac{1.5 ( 1. 1 ^{7} - 1 )}{1.1 - 1} = 15 (1.9487 - 1) \approx 14.2$ km.

We require $S_{n} > 100$ : $\frac{1.5 ( 1. 1 ^{n} - 1 )}{0.1} > 100$ , that is $15 (1. 1^{n} - 1) > 100$ .
So $1. 1^{n} > 1 + \frac{100}{15} = 7.667$ , and taking logarithms $n > \frac{ln 7.667}{ln 1.1} \approx 21.4$ . (Using a table on the GDC also works.)
Check: $S_{21} \approx 15 (7.4002 - 1) = 96.0$ km, which does not exceed 100, and $S_{22} \approx 15 (8.1403 - 1) = 107.1$ km, which does.
Therefore $n = 22$ .

Answer: a) $1.65$ km. b) $\approx 14.2$ km. c) He trains for $22$ days.

Exercise 46 MediumIB · P1 TZ0 November 2011

Consider the sequence $512, 128, 32, 8, \dots$

a) Calculate the exact value of the ninth term of the sequence.

b) Calculate the smallest number of terms needed for the sum of the sequence to be greater than $682.6$

Show answer

Answer: a) $u_{9} = \frac{1}{128}$ . b) $n = 7$ .

Show working and answer

The ratios are $\frac{128}{512} = \frac{32}{128} = \frac{8}{32} = \frac{1}{4}$ : a geometric sequence with $u_{1} = 512$ and $r = \frac{1}{4}$ .

a) $u_{9} = 512 (\frac{1}{4})^{8} = \frac{2 ^{9}}{2 ^{16}} = \frac{1}{2 ^{7}} = \frac{1}{128}$

$S_{n} = \frac{512 ( 1 - 0.2 5 ^{n} )}{1 - 0.25} = \frac{2048}{3} (1 - 0.2 5^{n})$ . The sum tends to $\frac{2048}{3} \approx 682.67$ .
We require $\frac{2048}{3} (1 - 0.2 5^{n}) > 682.6$ , that is $0.2 5^{n} < 1 - \frac{2047.8}{2048} = \frac{1}{10240}$ .
This is equivalent to $4^{n} > 10240$ . Using logarithms, $n > \frac{ln 10240}{ln 4} \approx 6.66$ . (Or use a table on the GDC.)
Check: $S_{6} = \frac{2048}{3} (1 - \frac{1}{4096}) \approx 682.50 < 682.6$ and $S_{7} = \frac{2048}{3} (1 - \frac{1}{16384}) \approx 682.63 > 682.6$ .
The smallest number of terms is $n = 7$ .

Answer: a) $u_{9} = \frac{1}{128}$ . b) $n = 7$ .

Exercise 47 MediumIB · P1 TZ0 November 2010

Consider the geometric sequence $16, 8, a, 2, b, \dots$

a) Write down the common ratio.

b) Write down the value of

(i) $a$ ;

(ii) $b$ .

c) The sum of the first $n$ terms is equal to $31.9375$ . Find the value of $n$ .

Show answer

Answer: a) $r = \frac{1}{2}$ ; b) $a = 4$ , $b = 1$ ; c) $n = 9$ .

Show working and answer

a) The ratio is the quotient of two consecutive terms: $r = \frac{8}{16} = \frac{1}{2}$ .

b) (i) $a = 8 \cdot \frac{1}{2} = 4$ . (Check: $4 \cdot \frac{1}{2} = 2$ , which is the following term.)

(ii) $b = 2 \cdot \frac{1}{2} = 1$ .

c) We use the sum of the first $n$ terms, $S_{n} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ :

$S_{n} = \frac{16 ( 1 - ( \frac{1}{2} ) ^{n} )}{1 - \frac{1}{2}} = 32 (1 - (\frac{1}{2})^{n})$ .
We equate: $32 (1 - (\frac{1}{2})^{n}) = 31.9375$ .
$1 - (\frac{1}{2})^{n} = \frac{31.9375}{32} = 0.998046875$ , so $(\frac{1}{2})^{n} = 0.001953125 = \frac{1}{512}$ .
As $512 = 2^{9}$ , we get $n = 9$ (this can also be found by solving the equation on the GDC).

Answer: a) $r = \frac{1}{2}$ ; b) $a = 4$ , $b = 1$ ; c) $n = 9$ .

Exercise 48 MediumIB · P1 TZ1 May 2011

The 7th term, $u_{7}$ , of a geometric sequence is 108. The 8th term, $u_{8}$ , of the sequence is 36.

a) Write down the common ratio of the sequence.

b) Find $u_{1}$ .

The sum of the first $k$ terms of the sequence is 118,096.

c) Find the value of $k$ .

Show answer

Answer: a) $r = \frac{1}{3}$ ; b) $u_{1} = 78732$ ; c) $k = 10$ .

Show working and answer

a) $r = \frac{u _{8}}{u _{7}} = \frac{36}{108} = \frac{1}{3}$ .

b) $u_{7} = u_{1} r^{6}$ , so $108 = u_{1} (\frac{1}{3})^{6} = \frac{u _{1}}{729}$ and $u_{1} = 108 \cdot 729 = 78732$ .

c) We use $S_{k} = \frac{u _{1} ( 1 - r ^{k} )}{1 - r}$ :

$S_{k} = \frac{78732 ( 1 - ( \frac{1}{3} ) ^{k} )}{\frac{2}{3}} = 118098 (1 - (\frac{1}{3})^{k})$ .
We equate to $118096$ : $1 - (\frac{1}{3})^{k} = \frac{118096}{118098}$ .
$(\frac{1}{3})^{k} = \frac{2}{118098} = \frac{1}{59049} = \frac{1}{3 ^{10}}$ .
Therefore $k = 10$ .

Answer: a) $r = \frac{1}{3}$ ; b) $u_{1} = 78732$ ; c) $k = 10$ .

Exercise 49 MediumIB · P2 TZ2 May 2009

A geometric sequence has a second term of 12 and a fifth term of 324.

a) Calculate the value of the common ratio.

b) Calculate the tenth term of this sequence.

c) The $k$ th term is the first term that is greater than 2000. Find the value of $k$ .

Show answer

Answer: a) $r = 3$ ; b) $u_{10} = 78732$ ; c) $k = 7$ .

Show working and answer

a) $u_{5} = u_{2} r^{3}$ , so $324 = 12 r^{3}$ , $r^{3} = 27$ and $r = 3$ .

b) First $u_{1} = \frac{u _{2}}{r} = \frac{12}{3} = 4$ . Then $u_{10} = 4 \cdot 3^{9} = 4 \cdot 19683 = 78732$ .

c) We look for the smallest $k$ with $4 \cdot 3^{k - 1} > 2000$ :

$3^{k - 1} > 500$ .
As $3^{5} = 243 < 500$ and $3^{6} = 729 > 500$ , we need $k - 1 = 6$ (on the GDC this can be checked in a table of values).
$k = 7$ . Check: $u_{7} = 4 \cdot 729 = 2916 > 2000$ and $u_{6} = 4 \cdot 243 = 972 < 2000$ .

Answer: a) $r = 3$ ; b) $u_{10} = 78732$ ; c) $k = 7$ .

Exercise 50 MediumIB · P1 TZ0 November 2016

A hydraulic hammer is used to drive a metal pole vertically into the ground.

To do this, it strikes the top of the pole repeatedly. The distance the pole penetrates the ground with the $n$ th blow of the hammer is equal to $d_{n}$ .

The distances $d_{1}, d_{2}, d_{3}, \dots, d_{n}$ form a geometric sequence.

The distance the pole penetrates the ground with the first blow of the hammer, $d_{1}$ , is equal to 64 cm.

The distance the pole penetrates the ground with the second blow of the hammer, $d_{2}$ , is equal to 48 cm.

a) Find the common ratio of this sequence.

b) Find the distance the pole penetrates the ground with the eighth blow of the hammer.

c) Find the total depth to which the pole has penetrated the ground after 10 blows of the hammer.

Show answer

Answer: $r = 0.75$ ; $d_{8} \approx 8.54$ cm; total depth $\approx 242$ cm.

Show working and answer

In a geometric sequence $r = \frac{d _{2}}{d _{1}} = \frac{48}{64}$ .
$r = 0.75$ .

Answer a): $r = 0.75$ .

$d_{n} = 64 \cdot 0.7 5^{n - 1}$ .
$d_{8} = 64 \cdot 0.7 5^{7} \approx 64 \cdot 0.13348 = 8.543$ .

Answer b): $d_{8} \approx 8.54$ cm.

The total depth after 10 blows is the sum $S_{10}$ of the geometric sequence.
$S_{10} = \frac{64 ( 1 - 0.7 5 ^{10} )}{1 - 0.75} = 256 (1 - 0.7 5^{10})$ .
With $0.7 5^{10} \approx 0.05631$ : $S_{10} \approx 256 \cdot 0.94369 \approx 241.58$ cm.

Answer c): $\approx 242$ cm.

Answer: $r = 0.75$ ; $d_{8} \approx 8.54$ cm; total depth $\approx 242$ cm.

Exercise 51 MediumIB · P1 TZ0 November 2013

512 players start round 1 of a tennis tournament. In this round, each player plays one match against another player. The player who wins the match goes through to the next round (round 2), while the player who loses is eliminated from the tournament. The tournament continues in this way until there is a winner.

a) Find the number of players who play in round 6 of the tournament.

b) Find the total number of matches played in the tournament.

Show answer

Answer: a) 16 players. b) 511 matches.

Show working and answer

The number of players in each round forms a geometric sequence with $u_{1} = 512$ and $r = \frac{1}{2}$ (each match eliminates one of every two players).

$u_{n} = 512 \cdot (\frac{1}{2})^{n - 1}$ .
Round 6: $u_{6} = 512 \cdot (\frac{1}{2})^{5} = \frac{512}{32} = 16$ players.

In each round, half as many matches are played as there are players: $256, 128, 64, \dots, 1$ . These are $256 \cdot (\frac{1}{2})^{k - 1}$ with $k = 1, \dots, 9$ (the last one, the final, is $1$ match: $256/ 2^{8} = 1$ ).
Sum of 9 terms with first term 256 and ratio $\frac{1}{2}$ : $S_{9} = \frac{256 ( 1 - ( \frac{1}{2} ) ^{9} )}{1 - \frac{1}{2}} = 511$ .
Check: there are 512 players and only one remains, so 511 players are eliminated, one per match.

Answer: a) 16 players. b) 511 matches.

Exercise 52 MediumIB · P1 TZ1 May 2018

Matryoshka dolls, or Russian dolls, are dolls of similar design that open up and fit one inside the other. The largest set of this type of doll is a set of 51 pieces that was completed in 2003. The height of the largest piece in this set is 54 cm. The heights of successive smaller dolls are 94% of the height of the previous larger doll, as shown.

Five Russian dolls in a row, of decreasing height, on a horizontal line. A double vertical arrow to the left of the first marks its height h; another arrow next to the second marks its height 0.94h

a) Find the height of the smallest doll in this set.

b) Find the total height if all 51 dolls were placed on top of each other.

Show answer

Answer: a) $\approx 2.45$ cm. b) $\approx 862$ cm.

Show working and answer

The heights form a geometric sequence with $u_{1} = 54$ and $r = 0.94$ .

The smallest doll is number 51: $u_{51} = 54 \cdot 0.9 4^{50}$ .
$u_{51} \approx 2.45$ cm.

$S_{51} = \frac{54 ( 1 - 0.9 4 ^{51} )}{1 - 0.94}$ .
$S_{51} \approx 861.65$ cm, that is, $\approx 862$ cm.

Answer: a) $\approx 2.45$ cm. b) $\approx 862$ cm.

Exercise 53 MediumIB · P1 TZ1 May 2009

The annual fees paid to a school during the school years 2000, 2001 and 2002 increase in the manner of a geometric sequence. The following table shows the fee structure.

Year	Fee (USD)
2000	8000.00
2001	8320.00
2002	8652.80

a) Calculate the common ratio for the increasing sequence of fees.

In parts (b) and (c), give your answer correct to 2 decimal places.

The fees continue to increase in the same proportion.

b) Find the fees paid for 2006.

A student attends the school for eight years, starting in 2000.

c) Find the total fees paid over these eight years.

Show answer

Answer: a) $r = 1.04$ ; b) $10,122.55; c) $73,713.81

Show working and answer

a) In a geometric sequence the ratio is the quotient of two consecutive terms.

$r = \frac{8320}{8000} = 1.04$
Check with the next pair: $\frac{8652.80}{8320} = 1.04$ .

Answer a): $r = 1.04$

b) 2000 is the term $u_{1} = 8000$ , so 2006 is the term $u_{7}$ .

Formula: $u_{n} = u_{1} r^{n - 1}$ .
$u_{7} = 8000 \cdot 1.0 4^{6} = 8000 \cdot 1.2653 \dots = 10122.55$

Answer b): $10,122.55

c) Eight years are the first eight terms: sum of a geometric sequence.

Formula: $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1}$ .
$S_{8} = \frac{8000 ( 1.0 4 ^{8} - 1 )}{0.04} = \frac{8000 \cdot 0.368569}{0.04} = 73713.81$

Answer: a) $r = 1.04$ ; b) $10,122.55; c) $73,713.81

Exercise 54 MediumIB · P2 TZ0 May 2006

Annie is about to start work. She will earn $ 26,000 during her first year, and her annual salary will increase by 3% each year.

a) Calculate how much Annie will earn during her fifth year of work.

Annie spends $ 24,800 of what she earns in her first year of work. In the following years, the cost of living will increase by 5% per year due to inflation.

b) (i) Calculate in how many years Annie will spend more than she earns.

(ii) By how much will her expenses exceed her income in that year?

Show answer

Answer: a) $ 29,263.23; b) (i) in the 4th year; (ii) expenses exceed income by approximately $ 298.

Show working and answer

The salary follows a geometric sequence with first term $26000$ and common ratio $1.03$ : $u_{n} = 26000 \cdot 1.0 3^{n - 1}$ .
Fifth year: $u_{5} = 26000 \cdot 1.0 3^{4} = 29263.23 \dots$

Answer a): $ 29,263.23.

Expenses also form a geometric sequence: $g_{n} = 24800 \cdot 1.0 5^{n - 1}$ .
Expenses exceed income when $24800 \cdot 1.0 5^{n - 1} > 26000 \cdot 1.0 3^{n - 1}$ . We use the GDC (table or intersection of the two curves): they cross at $n \approx 3.46$ .
We check with whole-number values:
- Year 3: income $26000 \cdot 1.0 3^{2} = 27583.40$ ; expenses $24800 \cdot 1.0 5^{2} = 27342.00$ . She still earns more than she spends.
- Year 4: income $26000 \cdot 1.0 3^{3} = 28410.90$ ; expenses $24800 \cdot 1.0 5^{3} = 28709.10$ . Now she spends more.
(i) It is in her fourth year of work, that is, after 4 years.
(ii) The difference is $28709.10 - 28410.90 = 298.20 \approx 298$ .

Answer: a) $ 29,263.23; b) (i) in the 4th year; (ii) expenses exceed income by approximately $ 298.

Exercise 55 MediumIB · P1 TZ0 November 2004

Herman joined a tennis club in the year 2000. The membership fee was £ 1,200 per year. It increases by 3% each year.

a) Calculate the membership fee for the year 2002, giving your answer to the nearest pound sterling (£).

b) Calculate the total amount Herman will pay in fees if he joined the club in the year 2000 and remains a member for five years.

View solution

The fees form a geometric sequence with first term $1200$ (year 2000) and common ratio $1.03$ : $u_{n} = 1200 \cdot 1.0 3^{n - 1}$ .
The year 2002 is the third year, $n = 3$ : $u_{3} = 1200 \cdot 1.0 3^{2} = 1273.08$ .
To the nearest unit: $1273$ .

Answer a): £ 1,273.

Five years of fees: the sum of the first 5 terms of the geometric sequence, $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1}$ .
$S_{5} = \frac{1200 ( 1.0 3 ^{5} - 1 )}{0.03} = 6370.97 \dots$

Answer b): £ 6,370.97 in total (about £ 6,371).

Exercise 56 HardIB · P2 TZ0 May 2007

A geometric sequence $G_{1}$ has first term 1 and common ratio 3.

a) The sum of the first $n$ terms of $G_{1}$ is equal to 29,524. Find $n$ .

A second geometric sequence $G_{2}$ is of the form $1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \dots, 1$ ,

b) State the common ratio of $G_{2}$ .

c) Calculate the sum of the first 10 terms of $G_{2}$ .

d) Explain why the sum of the first 1000 terms of $G_{2}$ is the same as the sum of the first 10 terms, when the result is rounded to three significant figures.

e) Using the results obtained in parts (a) to (c) or by any other method, calculate the sum of the first 10 terms of the sequence $2, 3 \frac{1}{3}, 9 \frac{1}{9}, 27 \frac{1}{27}, \dots, 2$ .

Give your answer correct to one decimal place.

Show answer

Answer: $n = 10$ ; $r = \frac{1}{3}$ ; $S_{10} \approx 1.50$ ; $S_{10}$ of the new sequence $= 29525.5$ .

Show working and answer

Geometric sum: $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1} = \frac{3 ^{n} - 1}{2}$ .
$\frac{3 ^{n} - 1}{2} = 29524 \Rightarrow 3^{n} = 59049$ .
As $3^{10} = 59049$ , $n = 10$ . (It can also be solved using the GDC: $\frac{3 ^{n} - 1}{2} = 29524$ .)

Answer a): $n = 10$ .

b) The ratio is the quotient of two consecutive terms: $r = \frac{1/3}{1} = \frac{1}{3}$ .

Answer b): $r = \frac{1}{3}$ .

c) 1. $S_{10} = \frac{1 \cdot ( 1 - ( \frac{1}{3} ) ^{10} )}{1 - \frac{1}{3}} = \frac{3}{2} (1 - \frac{1}{59049})$ . 2. $S_{10} \approx 1.49997 \dots$ , which to 3 significant figures is $1.50$ .

Answer c): $S_{10} \approx 1.50$ .

d) 1. Each term of $G_{2}$ is $(\frac{1}{3})^{k - 1}$ ; from the tenth term onwards they are extremely small. 2. The terms from the 11th to the 1000th add up to less than $\frac{( 1/3 ) ^{10}}{1 - 1/3} \approx 0.000025$ , an amount that does not change the first three significant figures. 3. Moreover, $(\frac{1}{3})^{1000} \approx 0$ , so $S_{1000} = \frac{1 - ( \frac{1}{3} ) ^{1000}}{1 - \frac{1}{3}} \approx \frac{1}{2/3} = 1.5$ . 4. Thus $S_{10}$ and $S_{1000}$ are both $1.50$ to three significant figures.

Answer d): the terms added after the tenth are negligible, and both sums round to $1.50$ .

e) 1. The terms of the new sequence are $2 = 1 + 1$ , $3 \frac{1}{3} = 3 + \frac{1}{3}$ , $9 \frac{1}{9} = 9 + \frac{1}{9}$ , etc.; that is, the $k$ th term is $3^{k - 1} + (\frac{1}{3})^{k - 1}$ (the sum of the terms of $G_{1}$ and $G_{2}$ ). 2. Its sum of 10 terms is the sum of those of $G_{1}$ and $G_{2}$ : $29524 + 1.49997 \dots = 29525.49997 \dots$ 3. Rounding to one decimal place: $29525.5$ .

Answer e): $29525.5$ .

Answer: $n = 10$ ; $r = \frac{1}{3}$ ; $S_{10} \approx 1.50$ ; $S_{10}$ of the new sequence $= 29525.5$ .

Exercise 57 HardIB · P2 TZ1 May 2019

John buys a new bicycle for 880 US dollars (USD) and pays for it with a Canadian credit card. The credit card company charges John a 4.2% commission to convert this purchase into Canadian dollars (CAD). The exchange rate is 1 USD = 1.25 CAD.

a) Calculate, in CAD, the total amount that Juan pays for the bicycle.

John insures his bicycle with a US company. The insurance company produces the following table for the value of the bicycle during each year.

Year	Value of the bicycle (USD)
1st	880
2nd	704
3rd	563.20
…

The values of the bicycle form a geometric sequence.

b) Find the value of the bicycle during the fifth year. Give your answer correct to two decimal places.

c) Calculate, in years, when the value of the bicycle will be less than 50 USD.

During the first year, John pays 120 USD to insure his bicycle. Each year the amount he pays to insure his bicycle is reduced by 3.50 USD.

d) Calculate the total amount John has paid to insure his bicycle over the first 5 years.

John bought the bicycle in 2008.

e) Justify why John should not insure his bicycle in 2019.

View solution

Price in CAD: $880 \cdot 1.25 = 1100$ CAD.
The $4.2 %$ commission is added: $1100 \cdot 1.042 = 1146.20$ .

Answer a): John pays $1146.20$ CAD.

Common ratio: $r = \frac{704}{880} = 0.8$ (and $\frac{563.20}{704} = 0.8$ ).
General term: $u_{n} = 880 \cdot 0. 8^{n - 1}$ .
Fifth year: $u_{5} = 880 \cdot 0. 8^{4} = 880 \cdot 0.4096 = 360.448$ .

Answer b): $360.45$ USD.

We set up $880 \cdot 0. 8^{n - 1} < 50$ , that is, $0. 8^{n - 1} < 0.056818$ .
Solve (using logarithms or a table of values on the GDC). Using logarithms, since $ln 0.8 < 0$ the inequality reverses: $n - 1 > \frac{ln 0.056818}{ln 0.8} = 12.85.$
The first integer is $n - 1 = 13$ , so $n = 14$ . Check: $u_{13} = 880 \cdot 0. 8^{12} = 60.54 > 50$ and $u_{14} = 880 \cdot 0. 8^{13} = 48.43 < 50$ .

Answer c): the value is less than $50$ USD during the $14 .^{th}$ year, that is, once $13$ years have elapsed.

The insurance payments form an arithmetic sequence with $u_{1} = 120$ and $d = - 3.50$ .
Sum of the first $5$ terms: $S_{5} = \frac{5}{2} [2 \cdot 120 + (5 - 1) (- 3.5)] = 2.5 (240 - 14) = 565$ .

Answer d): $565$ USD.

If it was bought in 2008 (year 1), then 2019 is year $12$ .
Value of the bicycle: $u_{12} = 880 \cdot 0. 8^{11} = 75.59$ USD.
Insurance for that year: $120 - 3.5 \cdot 11 = 81.50$ USD.
The insurance costs more than the bicycle is worth, since $81.50 > 75.59$ .

Answer e): in 2019 the insurance premium ( $81.50$ USD) is greater than the value of the bicycle ( $75.59$ USD), so it is not worth insuring it.

Simple and compound interest

Exercise 58 BasicIB · P1 TZ0 November 2007

Two siblings, Adam and Ben, each inherit $ 6,500. Adam invests his money in a bond that pays simple interest at an annual rate of 5%. Ben invests his money in a bank that pays compound interest at an annual rate of 4.5%.

a) Calculate the value of Adam's investment at the end of 6 years.

b) Calculate the value of Ben's investment at the end of 6 years. Give your answer correct to 2 decimal places.

View solution

With simple interest, the interest is the same each year: $I = C \cdot r \cdot t = 6500 \cdot 0.05 \cdot 6 = 1950$ .
Final value $= 6500 + 1950 = 8450$ .

Answer a): $ 8,450.

With annual compound interest, $V = C (1 + \frac{r}{100})^{n}$ , with $C = 6500$ , $r = 4.5$ , $n = 6$ .
$V = 6500 \cdot 1.04 5^{6} = 8464.69 \dots$

Answer b): $ 8,464.69.

Exercise 59 BasicIB · P1 TZ0 November 2006

Andrew invests 20,000 Swiss francs in a bank that offers simple interest of 2% per year for 8 years.

a) Find the interest earned after these 8 years.

Philip invests 20,000 Swiss francs for 6 years in a bank at a nominal annual interest rate of 5%, compounded quarterly.

b) Find the total amount in Philip's account after these 6 years.

View solution

Simple interest: $I = C \cdot r \cdot t = 20000 \cdot 0.02 \cdot 8$ .
$I = 3200$ .

Answer a): 3200 Swiss francs.

Quarterly compound interest: there are $k = 4$ periods per year, so in $6$ years there are $n = 4 \cdot 6 = 24$ periods, and the rate per period is $5/4 = 1.25 %$ .
$V = 20000 (1 + \frac{5}{400})^{24} = 20000 \cdot 1.012 5^{24}$ .
$V = 26947.0 \dots$ (value given by the GDC with its financial functions: $N = 6$ , $I % = 5$ , $P V = - 20000$ , $P / Y = C / Y = 4$ ).

Answer b): 26,947 Swiss francs.

Exercise 60 MediumIB · P1 TZ1 May 2019

Devra invested $k$ US dollars (USD) in an account that pays a nominal annual interest rate of 3.1%, compounded monthly. After 6 years, she has 1,100 USD in the account.

a) Calculate the value of $k$ . Give your answer correct to two decimal places.

Devra then bought a computer that cost 1,100 USD and sold it 4 years later for 350 USD.

b) Find the rate at which the computer depreciated per year.

View solution

This is compound interest with $n = 12$ periods per year, nominal rate $r = 3.1 %$ and $6$ years, that is, $72$ monthly periods. We use $F V = k (1 + \frac{r}{100 n})^{n t} .$
Substitute: $1100 = k (1 + \frac{3.1}{1200})^{72}$ .
Using the GDC (financial solver) with $N = 6$ , $I % = 3.1$ , $F V = 1100$ , $P / Y = C / Y = 12$ , we obtain $P V$ (in absolute value); or alternatively, rearrange: $k = \frac{1100}{( 1 + \frac{3.1}{1200} ) ^{72}} = \frac{1100}{1.20413 \dots} = 913.52.$

Answer a): $k = 913.52$ USD.

Constant annual depreciation as a percentage: $V = V_{0} (1 - \frac{r}{100})^{t}$ .
With $V_{0} = 1100$ , $V = 350$ and $t = 4$ : $350 = 1100 (1 - \frac{r}{100})^{4}$ .
Rearranging: $(1 - \frac{r}{100})^{4} = \frac{350}{1100} = 0.31818 \dots$
Take the fourth root: $1 - \frac{r}{100} = 0.3181 8^{1/4} = 0.75105$ .
Therefore $\frac{r}{100} = 0.24895$ , that is, $r \approx 24.9 %$ .

Answer b): the computer depreciated at a rate of approximately $24.9 %$ per year.

Exercise 61 MediumIB · P1 TZ0 November 2019

Siân invests 50,000 Australian dollars (AUD) in a savings account which offers a nominal annual interest rate of 5.6% compounded monthly.

a) Calculate the value of Siân's investment after four years. Give your answer correct to two decimal places.

After that four-year period, Siân withdraws 40,000 AUD from her savings account and uses this money to buy a car. We know that the car depreciates by 18% each year. After $t$ years, the car will be worth 2,500 AUD.

b) Find the value of $t$ .

View solution

Monthly compound interest: $n = 12$ , rate $5.6 %$ , $t = 4$ years, that is, $48$ periods.
$F V = 50000 (1 + \frac{5.6}{1200})^{48}$ .
Using the GDC (or a calculator): $F V = 50000 \cdot 1.250419 \dots = 62520.97$ .

Answer a): $62 520.97$ AUD.

The car is worth $40000$ AUD when bought and loses $18 %$ per year, so each year it is multiplied by $0.82$ : $40000 \cdot 0.8 2^{t} = 2500.$
Divide both sides: $0.8 2^{t} = \frac{2500}{40000} = 0.0625$ .
Take logarithms (or solve the equation using the GDC): $t = \frac{ln 0.0625}{ln 0.82} = \frac{- 2.77259}{- 0.19845} = 13.97.$

Answer b): $t \approx 13.97$ years (approximately $14$ years).

Exercise 62 MediumIB · P1 TZ2 May 2018

In this question, give all answers correct to two decimal places.

Karl invests 1,000 US dollars (USD) in an account that pays a nominal annual interest rate of $3.5 %$ , compounded quarterly. Karl leaves the money in this account for 5 years.

a) (i) Calculate the amount of money Karl will have in the account after 5 years;

(ii) Write down the amount of interest he will have received after those 5 years.

Karl decides to donate this interest to an NGO in France. The NGO receives 170 euros (EUR). The exchange rate is $1 USD = t EUR$ .

b) Calculate the value of $t$ .

View solution

a) (i)

Quarterly compound interest: $n = 4$ , $t = 5$ years, $20$ periods, rate $3.5 %$ .
$F V = 1000 (1 + \frac{3.5}{400})^{20} = 1000 \cdot 1.19034 \dots = 1190.34$ .

(ii)

Interest $=$ final amount $-$ initial principal $= 1190.34 - 1000 = 190.34$ .

Answer a): (i) $1190.34$ USD; (ii) $190.34$ USD.

We have $190.34 \cdot t = 170$ .
$t = \frac{170}{190.34} = 0.893 \dots \approx 0.89$ .

Answer b): $t = 0.89$ EUR per USD.

Exercise 63 MediumIB · P1 TZ0 November 2018

Nick has 150,000 $ in a trust. Every year, he donates $8 %$ of the money remaining in the trust.

a) Determine the maximum number of years for which Nick can donate money, while keeping at least 50,000 $ in the trust.

Louise invests 200,000 $ in a bank account that pays a nominal interest rate of $5 %$ , compounded quarterly, for eight years.

b) Calculate the value of Louise's investment when this period ends. Give your answer correct to the nearest cent.

View solution

Each year $92 %$ of the money remains, so after $n$ years there is $150000 \cdot 0.9 2^{n}$ .
We require $150000 \cdot 0.9 2^{n} \geq 50000$ , that is, $0.9 2^{n} \geq \frac{1}{3}$ .
Using logarithms (the inequality reverses because $ln 0.92 < 0$ ): $n \leq \frac{ln ( 1/3 )}{ln 0.92} = \frac{- 1.0986}{- 0.08338} = 13.18.$
Check: $150000 \cdot 0.9 2^{13} = 50760 > 50000$ and $150000 \cdot 0.9 2^{14} = 46700 < 50000$ .

Answer a): he can donate for a maximum of $13$ years.

Quarterly compound interest: $n = 4$ , $8$ years, $32$ periods, rate $5 %$ .
$F V = 200000 (1 + \frac{5}{400})^{32} = 200000 \cdot 1.012 5^{32} = 200000 \cdot 1.488131 \dots = 297 626.10$ .

Answer b): $297 626.10$ $.

Exercise 64 MediumIB · P1 TZ1 May 2017

Arthur and Jacob dream of owning a speedboat that costs 35,300 euros (EUR). Arthur invested $x$ EUR in an account that pays a nominal annual interest rate of $3.6%$ , compounded monthly. After 18 years, he will have 35,300 EUR in the account.

a) Calculate the value of Arthur's initial investment, $x$ . Give your answer correct to two decimal places.

Jacob invested 9,000 EUR for $n$ years. The investment has a nominal annual interest rate of $3.2%$ and is compounded quarterly. After $n$ years, the investment will be worth 35,300 EUR.

b) Find the value of $n$ .

Show answer

Answer: a) $x = 18483.03$ EUR; b) $n \approx 42.9$ years (approx. $43$ years).

Show working and answer

a) Monthly compound interest: $F V = P V (1 + \frac{r}{100 k})^{k n}$ with $F V = 35300$ , $r = 3.6$ , $k = 12$ , $n = 18$ .

We write the equation: $35300 = x (1 + \frac{3.6}{1200})^{12 \cdot 18} = x (1.003)^{216}$ .
We solve for $x$ (or ask the GDC's finance tool: $N = 216$ , $I % = 3.6$ , $F V = 35300$ , $P V = ?$ , $P / Y = C / Y = 12$ ): $x = \frac{35300}{1.00 3 ^{216}} \approx 18483.03$ .

Answer a): $x = 18483.03$ EUR.

b) Quarterly compound interest ( $k = 4$ , rate per quarter $\frac{3.2}{4} = 0.8%$ ):

We set up $9000 (1.008)^{4 n} = 35300$ .
Divide: $(1.008)^{4 n} = \frac{35300}{9000} \approx 3.9222$ .
Solve using the GDC (equation solver or the finance tool with $P V = - 9000$ , $F V = 35300$ , $I % = 3.2$ , $P / Y = C / Y = 4$ ) or with logarithms: $4 n = \frac{ln 3.9222}{ln 1.008} \approx 171.5$ .
Therefore $n \approx 42.9$ years, that is, about $43$ years.

Answer: a) $x = 18483.03$ EUR; b) $n \approx 42.9$ years (approx. $43$ years).

Exercise 65 MediumIB · P1 TZ0 November 2017

Juan buys a bicycle in the sales. Juan receives a $30 %$ discount on the original price and pays 560 US dollars (USD) for the bicycle.

a) Calculate the original price of the bicycle.

In order to buy the bicycle, Juan takes out a loan of 560 USD for 6 months with a nominal annual interest rate of $75 %$ , compounded monthly. Juan believes that the total amount he will have to pay will be less than the original price of the bicycle.

b) Calculate the difference between the original price of the bicycle and the total amount that Juan will end up paying.

Show answer

Answer: a) $800$ USD; b) Juan will pay about $5.68$ USD (approx. $6$ USD) more than the original price.

Show working and answer

a) The price paid is $70%$ of the original (a $30%$ discount).

$0.70 \cdot P = 560$ .
$P = \frac{560}{0.7} = 800$ .

Answer a): the original price was $800$ USD.

b) Monthly compound interest: monthly rate $\frac{75}{12} = 6.25%$ , for $6$ months.

Amount to pay: $560 (1 + 0.0625)^{6} = 560 \cdot 1.062 5^{6} \approx 805.68$ USD.
Comparing with the original price: $805.68$ is greater than $800$ , so Juan was wrong: he will pay more than the original price.
Difference: $805.68 - 800 = 5.68$ USD, that is, about $6$ USD more.

Answer: a) $800$ USD; b) Juan will pay about $5.68$ USD (approx. $6$ USD) more than the original price.

Exercise 66 MediumIB · P1 TZ1 May 2016

In this question, give all correct answers to two decimal places.

Diogo deposited 8,000 Argentine pesos, ARS, in a bank account that pays a nominal annual interest rate of $15%$ , compounded monthly.

a) Find how much interest Diogo has earned after 2 years.

Carmen also deposited ARS in a bank account. Her account pays a nominal annual interest rate of $17%$ , compounded annually. After three years, the total amount in Carmen's account is 10,000 ARS.

b) Find the amount that Carmen deposited in the bank account.

Show answer

Answer: a) $2778.81$ ARS; b) $6243.71$ ARS.

Show working and answer

a) Monthly compound interest: $k = 12$ , monthly rate $\frac{15}{12} = 1.25%$ , $n = 2$ years, that is, $24$ periods.

Final value: $8000 (1.0125)^{24} \approx 10778.81$ ARS.
Interest earned = final value $-$ deposit: $10778.81 - 8000 = 2778.81$ .

Answer a): $2778.81$ ARS of interest.

b) Annual compound interest at $17%$ for $3$ years.

$10000 = C (1.17)^{3}$ .
$C = \frac{10000}{1.1 7 ^{3}} = \frac{10000}{1.601613} \approx 6243.71$ .

Answer: a) $2778.81$ ARS; b) $6243.71$ ARS.

Exercise 67 MediumIB · P1 TZ2 May 2016

Obi travels from Dubai to Pretoria and, for this, exchanges 2,000 United Arab Emirates dirhams (AED) at a bank. In return he receives 6,160 South African rand (ZAR). The exchange rate applied is $1 AED = x ZAR$ .

a) Calculate the value of $x$ .

Obi decides to invest half of the money he has received, that is, 3,080 ZAR, in an account that pays a nominal interest rate of $9 %$ , compounded monthly. The amount of money in the account will have doubled before the end of the $n$ th year of investment.

b) Calculate the minimum value of $n$ .

Show answer

Answer: a) $x = 3.08$ ; b) $n = 8$ (it doubles before the end of the eighth year).

Show working and answer

a) The exchange rate is the amount of ZAR received for each AED.

$x = \frac{6160}{2000} = 3.08$ .

Answer a): $x = 3.08$ ZAR per AED.

b) Monthly rate $\frac{9}{12} = 0.75%$ . With $m$ months of investment:

Doubling condition: $3080 (1.0075)^{m} = 6160$ , that is, $(1.0075)^{m} = 2$ .
Using the GDC (equation solver or table of values) or logarithms: $m = \frac{ln 2}{ln 1.0075} \approx 92.76$ months.
The money doubles between months $92$ and $93$ . As $92.76$ months is $7.73$ years, the doubling occurs during the eighth year.
Check: at the end of year $7$ (month $84$ ) there is $3080 \cdot 1.007 5^{84} \approx 5772.6 < 6160$ ; at the end of year $8$ (month $96$ ) there is $3080 \cdot 1.007 5^{96} \approx 6309.7 > 6160$ .

Answer: a) $x = 3.08$ ; b) $n = 8$ (it doubles before the end of the eighth year).

Exercise 68 MediumIB · P1 TZ1 May 2015

Pierre invests 5,000 euros in a fixed deposit that pays a nominal annual interest rate of $4.5%$ , compounded monthly, for seven years.

a) Calculate the value of Pierre's investment at the end of this time. Give your answer correct to two decimal places.

Carla has 7,000 dollars to invest in a fixed deposit that is compounded annually. She wants to double her money after 10 years.

b) Calculate the minimum annual interest rate needed for Carla to achieve her goal.

Show answer

Answer: a) $6847.26$ euros; b) minimum rate $r \approx 7.18%$ per year.

Show working and answer

a) Monthly compound interest: monthly rate $\frac{4.5}{12} = 0.375%$ , $7 \cdot 12 = 84$ periods.

$5000 (1.00375)^{84} \approx 6847.26$ .

Answer a): $6847.26$ euros.

b) Annual compound interest; we want $7000 \to 14000$ in $10$ years.

$7000 (1 + \frac{r}{100})^{10} = 14000$ .
$(1 + \frac{r}{100})^{10} = 2$ , so $1 + \frac{r}{100} = 2^{1/10} \approx 1.0718$ .
(It can also be solved using the GDC.) Hence $r \approx 7.18$ .

Answer: a) $6847.26$ euros; b) minimum rate $r \approx 7.18%$ per year.

Exercise 69 MediumIB · P1 TZ2 May 2015

Minta deposits 1,000 euros in a bank account. The bank pays a nominal annual interest rate of $5%$ , compounded quarterly.

a) Calculate the amount of money Minta will have in her account after 3 years. Give your answer correct to two decimal places.

Minta will withdraw the money from her bank account when the interest accrued is 300 euros.

b) Calculate the time, in years, until Minta withdraws the money from her bank account.

Show answer

Answer: a) $1160.75$ euros; b) $t \approx 5.28$ years.

Show working and answer

a) Quarterly compound interest: rate per quarter $\frac{5}{4} = 1.25%$ , $3 \cdot 4 = 12$ quarters.

$1000 (1.0125)^{12} \approx 1160.75$ .

Answer a): $1160.75$ euros.

b) The interest is $300$ euros when the balance is $1300$ euros.

$1000 (1.0125)^{4 t} = 1300$ , so $(1.0125)^{4 t} = 1.3$ .
Using the GDC (equation or finance tool) or logarithms: $4 t = \frac{ln 1.3}{ln 1.0125} \approx 21.12$ quarters.
$t = \frac{21.12}{4} \approx 5.28$ years.

Answer: a) $1160.75$ euros; b) $t \approx 5.28$ years.

Exercise 70 MediumIB · P1 TZ1 May 2014

Ludmila takes out a loan of 320,000 Brazilian reals (BRL) from a bank for two years at a nominal annual interest rate of $10%$ , compounded semi-annually.

a) Write down the number of times interest is added to the loan over the two years.

b) Calculate the exact amount of money that Ludmila must repay at the end of the two years.

Ludmila estimates that she will have to repay 360,000 BRL at the end of the two years.

c) Calculate the percentage error in her estimate.

Show answer

Answer: a) $4$ times; b) $388962$ BRL; c) error of $7.45%$ .

Show working and answer

a) Semi-annual compounding: $2$ times a year.

In $2$ years: $2 \cdot 2 = 4$ times.

Answer a): $4$ times.

b) Rate per half-year: $\frac{10}{2} = 5%$ , with $4$ periods.

$320000 (1.05)^{4} = 320000 \cdot 1.21550625 = 388962$ .

Answer b): $388962$ BRL.

c) Percentage error $= \frac{v _{estimated} - v _{exact}}{v _{exact}} \cdot 100$ .

$\frac{360000 - 388962}{388962} \cdot 100 = \frac{28962}{388962} \cdot 100 \approx 7.45$ .

Answer: a) $4$ times; b) $388962$ BRL; c) error of $7.45%$ .

Exercise 71 MediumIB · P1 TZ0 November 2014

Mandzur, a farmer, applies for a loan to buy a buffalo. He borrows 900,000 Cambodian riel (KHR) for 2 years. The nominal annual interest rate is $15%$ , compounded monthly.

a) Find the total interest Mandzur must pay. Give your answer correct to the nearest 100 KHR.

b) Write your answer to part (a) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10, k \in Z$ .

Show answer

Answer: a) $312600$ KHR; b) $3.126 \times 1 0^{5}$ .

Show working and answer

a) Monthly rate $\frac{15}{12} = 1.25%$ , $24$ periods.

Final value: $900000 (1.0125)^{24} \approx 1212615.9$ KHR.
Interest: $1212615.9 - 900000 \approx 312615.9$ .
Rounding to the nearest $100$ KHR: $312600$ .

Answer a): $312600$ KHR.

b) Scientific notation:

$312600 = 3.126 \times 1 0^{5}$ .

Answer: a) $312600$ KHR; b) $3.126 \times 1 0^{5}$ .

Exercise 72 MediumIB · P1 TZ1 May 2013

Jenny invested $ 20,000 in a bank account paying simple annual interest of $3.5%$ . She withdrew her investment from the account when its value was $ 31,200.

a) Find the number of years for which the money was invested.

Ramón invests $ 18,000 in a bank account that pays $3.4%$ nominal annual interest, compounded quarterly.

b) Find the minimum number of years Ramón must invest the money for his investment to reach a value of $ 27,000.

Show answer

Answer: a) $16$ years; b) $12$ years at minimum.

Show working and answer

a) Simple interest: each year the same amount is earned, $I = P \cdot r$ .

Annual interest: $20000 \cdot 0.035 = 700$ $ per year.
Total interest earned: $31200 - 20000 = 11200$ $.
Number of years: $n = \frac{11200}{700} = 16$ .

Answer a): $16$ years.

b) Compound interest: $F V = P V (1 + \frac{r}{100 k})^{k n}$ with $k = 4$ payments per year, $r = 3.4$ .

We set up $18000 (1 + \frac{3.4}{400})^{4 n} = 27000$ , that is, $18000 \cdot 1.008 5^{4 n} = 27000$ .
Using the GDC (equation solver, or the TVM finance function with $N$ unknown) we obtain $4 n \approx 47.9$ , that is, $n \approx 11.98$ years.
As $n$ must be such that the investment reaches $ 27,000 and the interest is paid each quarter, $48$ quarters are needed, that is, $12$ years. (With $11.75$ years, 47 quarters, it has not yet reached this value.)

Answer: a) $16$ years; b) $12$ years at minimum.

Exercise 73 MediumIB · P1 TZ2 May 2013

Marcus's grandmother gives him 500 Australian dollars (AUD) for his 18th birthday. He plans to deposit this money in a bank that offers a nominal annual interest rate of $6.0 %$ compounded quarterly for three years.

a) Calculate the total interest, in AUD, that Marcus would earn over these three years. Give your answer correct to two decimal places.

Marcus would earn the same interest over those three years if he deposited the 500 AUD in a second bank that offers interest compounded annually.

b) Calculate the interest rate offered by this second bank.

Show answer

Answer: a) $97.81$ AUD; b) $r = 6.14%$ per annum.

Show working and answer

a) Final amount with quarterly compound interest: $k = 4$ , $n = 3$ , $r = 6%$ .

$F V = 500 (1 + \frac{6}{400})^{12} = 500 \cdot 1.01 5^{12}$ .
Using the GDC: $F V = 597.81$ AUD (approx., $597.809 \dots$ ).
Interest: $597.81 - 500 = 97.81$ AUD.

Answer a): $97.81$ AUD.

b) At the second bank the final amount is the same, but the interest is compounded once a year for 3 years.

We set up $500 (1 + \frac{r}{100})^{3} = 597.809 \dots$
We divide by 500: $(1 + \frac{r}{100})^{3} = 1.195618 \dots$
Cube root: $1 + \frac{r}{100} = 1.0614$ , so $r \approx 6.14$ . (It can also be solved using the GDC, with $N = 3$ , $I %$ unknown and 1 payment per year.)

Answer: a) $97.81$ AUD; b) $r = 6.14%$ per annum.

Exercise 74 MediumIB · P1 TZ1 May 2012

Veronica wants to make an investment and accumulate EUR 25,000 over a period of 18 years. She finds two investment options. Option 1 offers simple interest of $5%$ per annum.

a) Find the exact amount she will have in her account after 18 years if she invests EUR 12,500 with this option.

Option 2 offers a nominal annual interest rate of $4%$ , compounded monthly.

b) Find the amount Verónica must invest with option 2 to have EUR 25,000 in her account after 18 years. Give your answer correct to two decimal places.

Show answer

Answer: a) $23750$ EUR; b) $12183.39$ EUR.

Show working and answer

a) Simple interest: $F V = P V (1 + \frac{r n}{100})$ .

Annual interest: $12500 \cdot 0.05 = 625$ EUR.
After 18 years: $625 \cdot 18 = 11250$ EUR.
Final amount: $12500 + 11250 = 23750$ EUR.

Answer a): $23750$ EUR.

b) Here we look for the initial capital $P V$ : compound monthly interest, $k = 12$ , $n = 18$ , $r = 4%$ .

Number of periods: $12 \cdot 18 = 216$ .
Equation: $P V (1 + \frac{4}{1200})^{216} = 25000$ .
We isolate $P V$ : $P V = \frac{25000}{( 1 + \frac{1}{300} ) ^{216}}$ .
Using the GDC (TVM with $P V$ unknown, or calculating directly): $P V \approx 12183.39$ EUR.

Answer: a) $23750$ EUR; b) $12183.39$ EUR.

Exercise 75 MediumIB · P1 TZ0 November 2012

Jackson invested AUD 12,000 in a bank that offered simple interest at an annual interest rate of $r %$ .

The value of Jackson's investment doubled after 20 years.

a) Calculate the value of $r$ .

Maddison invests AUD 15,000 in a bank that offers compound interest at an annual interest rate of 4.44%, compounded quarterly.

b) Calculate the number of years it will take for Maddison's investment to triple in value.

Show answer

Answer: a) $r = 5$ ; b) $25$ years (approx. $24.9$ years).

Show working and answer

a) With simple interest, doubling the value means earning $12000$ AUD of interest in 20 years.

Annual interest: $\frac{12000}{20} = 600$ AUD.
Rate: $r = \frac{600}{12000} \cdot 100 = 5$ .

Answer a): $r = 5$ .

b) Compound interest, quarterly ( $k = 4$ ), $r = 4.44%$ . Tripling means reaching $45000$ AUD.

Equation: $15000 (1 + \frac{4.44}{400})^{4 n} = 45000$ , that is, $1.011 1^{4 n} = 3$ .
Using the GDC (TVM with $N$ unknown, or solving the equation) we obtain $4 n \approx 99.5$ quarters, that is, $n \approx 24.9$ years.
As interest is paid at the end of each quarter, $100$ complete quarters are needed, that is, $25$ years (after $99$ quarters it has not yet tripled).

Answer: a) $r = 5$ ; b) $25$ years (approx. $24.9$ years).

Exercise 76 MediumIB · P1 TZ0 November 2011

Manisha borrows 200,000 Indian rupees (INR) from a lender for 18 months at an annual rate of 15% simple interest.

a) Calculate the interest that Manisha must pay at the end of the 18 months.

Kunal borrows 200,000 INR from the lender for 18 months at an annual interest rate of 15%, compounded monthly.

b) Calculate the total amount that Kunal must repay at the end of the 18 months.

Give your answer to the nearest rupee.

Show answer

Answer: a) $45000$ INR; b) $250115$ INR.

Show working and answer

a) Simple interest over $18$ months $= 1.5$ years.

$I = 200000 \cdot 0.15 \cdot 1.5 = 45000$ .

Answer a): $45000$ INR.

b) Compound monthly interest: $k = 12$ , $n = 1.5$ , number of periods $12 \cdot 1.5 = 18$ .

$F V = 200000 (1 + \frac{15}{1200})^{18} = 200000 \cdot 1.012 5^{18}$ .
Using the GDC: $F V \approx 250115.5 \dots$ , which to the nearest rupee is $250115$ .

Answer: a) $45000$ INR; b) $250115$ INR.

Exercise 77 MediumIB · P1 TZ2 May 2010

Astrid invests 1200 euros for five years, at a nominal annual interest rate of 7.2%, compounded monthly.

a) Find the interest earned by Astrid over the five years of the investment. Give your answer correct to two decimal places.

Helen invests 1200 euros for five years, at a simple annual interest rate. In total, she earns the same interest as Astrid.

b) Find the simple annual interest rate applied to Helen.

Show answer

Answer: a) $518.15$ EUR; b) $8.64%$ per annum.

Show working and answer

a) Compound monthly interest: $k = 12$ , $5$ years $\Rightarrow 60$ periods, monthly rate $\frac{7.2}{12} = 0.6%$ .

$F V = 1200 \cdot 1.00 6^{60} \approx 1718.15$ EUR.
Interest: $1718.15 - 1200 = 518.15$ EUR.

Answer a): $518.15$ EUR.

b) With simple interest, $I = P \cdot \frac{r}{100} \cdot n$ .

$518.15 = 1200 \cdot \frac{r}{100} \cdot 5 = 60 r$ .
$r = \frac{518.15}{60} = 8.64$ .

Answer: a) $518.15$ EUR; b) $8.64%$ per annum.

Exercise 78 MediumIB · P1 TZ2 May 2009

Inge borrows 4,500 € for 2 years.

a) Bank 1 charges compound interest at an annual rate of 15%, compounded quarterly.

Calculate the total amount to be repaid at the end of the 2 years.

Give your answer correct to two decimal places.

b) Bank 2 charges simple interest, and Inge would pay the same total amount as for Bank 1.

Find the interest rate charged by Bank 2.

Show answer

Answer: a) 6041.12 €; b) 17.1% per annum (simple)

Show working and answer

a) Quarterly compounding: $k = 4$ , $n = 2$ years, rate per quarter $\frac{15}{4} = 3.75 %$ .

$F V = 4500 (1 + \frac{15}{400})^{8} = 4500 \cdot 1.037 5^{8}$
$F V = 4500 \cdot 1.34247 \dots = 6041.12$

Answer a): 6041.12 €

b) With simple interest the same amount is paid, so the interest is $6041.12 - 4500 = 1541.12$ .

$1541.12 = 4500 \cdot \frac{r}{100} \cdot 2$
$r = \frac{1541.12}{90} = 17.12$

Answer: a) 6041.12 €; b) 17.1% per annum (simple)

Exercise 79 MediumIB · P1 TZ2 May 2008

Emma deposits 8,000 € in a bank account that pays a nominal annual interest rate of 5%, compounded quarterly.

a) Calculate the amount of money Emma would have in her account after 15 years.

Give your answer rounded to the nearest euro.

b) After some time she decides to withdraw the money from this bank. She has 9,058.17 € in her account. Calculate the number of months for which Emma had left the money in the account.

Show answer

Answer: a) 16857 €; b) 30 months

Show working and answer

a) $k = 4$ , $n = 15$ , rate per quarter $\frac{5}{4} = 1.25 %$ .

$F V = 8000 (1 + \frac{5}{400})^{60} = 8000 \cdot 1.012 5^{60}$
$F V = 8000 \cdot 2.10718 \dots = 16857.46$

Answer a): 16857 €

b) 1. $8000 \cdot 1.012 5^{m} = 9058.17$ , where $m$ is the number of quarters. 2. Using the GDC (solving the equation): $m = 10$ . (Check: $8000 \cdot 1.012 5^{10} = 9058.17$ .) 3. Months: $10 \cdot 3 = 30$ .

Answer: a) 16857 €; b) 30 months

Exercise 80 MediumIB · P1 TZ0 November 2008

Eva invests USD 2,000 at a nominal annual interest rate of 8%, compounded semi-annually.

a) Calculate the value of the investment after 5 years, rounded to the nearest whole dollar.

Tony invests USD 1,500 at a nominal annual interest rate of 7.8%, compounded annually.

b) Find the number of complete years it will take for Tony's investment to double in value.

Show answer

Answer: a) 2960 USD; b) 10 years

Show working and answer

a) $k = 2$ , $n = 5$ , semi-annual rate $\frac{8}{2} = 4 %$ .

$F V = 2000 (1 + \frac{8}{200})^{10} = 2000 \cdot 1.0 4^{10}$
$F V = 2000 \cdot 1.48024 = 2960.49$

Answer a): 2960 USD

b) Doubling: $1500 \cdot 1.07 8^{n} > 3000$ , that is, $1.07 8^{n} > 2$ .

Using the GDC, $1.07 8^{n} = 2$ gives $n \approx 9.23$ .
As complete years are requested and the doubled value must be exceeded: $n = 10$ . (With $n = 9$ the value is 2,959.59 USD $< 3000$ ; with $n = 10$ it is 3,190.4 USD.)

Answer: a) 2960 USD; b) 10 years

Exercise 81 MediumIB · P2 TZ0 May 2006

Alex invests 3,600 euros in an account that pays a nominal annual interest rate of 5.4%, compounded monthly. The interest is added to the account at the end of each month.

a) Calculate the number of whole months it will take for Alex's investment to double.

b) (i) Calculate the value of the investment after nine years.

(ii) Find the simple annual interest rate that would give the same value for the investment after nine years.

Show answer

Answer: a) 155 months; b) (i) 5846.50 €; (ii) 6.93%

Show working and answer

a) Monthly rate: $\frac{5.4}{12} = 0.45 %$ .

We seek $n$ with $3600 \cdot 1.004 5^{n} > 7200$ , that is, $1.004 5^{n} > 2$ .
Using the GDC, $1.004 5^{n} = 2$ gives $n \approx 154.4$ .
With $n = 154$ it has not yet doubled; with $n = 155$ it has.

Answer a): 155 months

b) (i) Nine years are $12 \cdot 9 = 108$ months.

$F V = 3600 \cdot 1.004 5^{108} = 3600 \cdot 1.62403 = 5846.50$

Answer (i): 5846.50 €

(ii) The interest earned is $5846.50 - 3600 = 2246.50$ .

Simple interest: $2246.50 = 3600 \cdot \frac{r}{100} \cdot 9$
$r = \frac{2246.50}{324} = 6.93$

Answer (ii): 6.93% per annum

Answer: a) 155 months; b) (i) 5846.50 €; (ii) 6.93%

Exercise 82 MediumIB · P1 TZ0 May 2007

a) 1 Brazilian real (BRL) = 2.607 South African rand (ZAR). Giving your answers correct to two decimal places,

(i) convert 300 BRL to ZAR,

(ii) find how many reais are needed to buy 300 rand.

Marta deposits in a savings account a gift from her uncle worth 150 reais. The savings account pays a simple annual interest rate of $r$ % per annum. The interest is added to the account at the end of each month. After 9 months, the amount in the savings account rose to 158.10 reais.

b) Find the value of $r$ .

Show answer

Answer: a) (i) 782.10 ZAR; (ii) 115.07 BRL; b) $r = 7.2$

Show working and answer

a) (i) 1. $300 \cdot 2.607 = 782.10$

Answer (i): 782.10 ZAR

(ii) 1. Divide by the exchange rate: $\frac{300}{2.607} = 115.07$

Answer (ii): 115.07 BRL

b) 1. Interest earned: $158.10 - 150 = 8.10$ reais. 2. 9 months are $\frac{9}{12} = 0.75$ years. 3. Simple interest: $8.10 = 150 \cdot \frac{r}{100} \cdot 0.75$ 4. $r = \frac{8.10}{1.125} = 7.2$

Answer: a) (i) 782.10 ZAR; (ii) 115.07 BRL; b) $r = 7.2$

Exercise 83 MediumIB · P1 TZ0 May 2005

A family in Malaysia receives a gift of 4,000 Australian dollars (AUD) from a cousin who lives in Australia. The money is converted into Malaysian ringgit. One ringgit is worth 0.4504 AUD.

a) Calculate the amount of ringgit received.

The money is invested for 2 years and 6 months at an annual rate of 5.2%, compounded monthly.

b) Calculate the amount of interest earned on this investment. Give your answer to the nearest ringgit.

View solution

If $1$ ringgit $= 0.4504$ AUD, to convert from AUD to ringgit we divide by $0.4504$ .
$4000/0.4504 = 8880.99 \dots$

Answer a): 8,880.99 ringgit.

Monthly compound interest: $k = 12$ , time $2.5$ years, so $n = 12 \cdot 2.5 = 30$ periods, with a monthly rate of $5.2/12 %$ .
Final amount: $V = 8880.99 (1 + \frac{5.2}{1200})^{30} = 10111.06 \dots$
Interest $= 10111.06 - 8880.99 = 1230.07 \dots$
Rounding to the nearest unit: $1230$ .

Answer b): 1230 ringgit of interest.

Exercise 84 MediumIB · P1 TZ0 November 2004

Bobby leaves America to travel for a year through France and England. Consider the following exchange rates:

1 US dollar (USD) = 0.983 Euros

1 British pound sterling (GBP) = 1.59 Euros

a) Bobby exchanges 500 USD into Euros.

(i) Calculate how many Euros he receives.

He spends 328 Euros in France and exchanges the rest into GBP.

(ii) Calculate how many GBP he receives.

While in England, Bobby decides to deposit this money in a bank that pays 6% simple interest per year, and he gets a part-time job to cover his expenses. Bobby stays in England for six months.

b) Calculate the interest he earns after the six months.

Show answer

Answer: a) (i) 491.5 euros; (ii) $\approx$ 103 GBP; b) about 3.08 GBP of interest (3.09 using 103 GBP).

Show working and answer

(i) To convert USD to euros, we multiply by $0.983$ : $500 \cdot 0.983 = 491.5$ .
(ii) After spending 328 euros, $491.5 - 328 = 163.5$ euros remain.
As $1$ GBP $= 1.59$ euros, to convert to pounds we divide by $1.59$ : $163.5/1.59 = 102.83 \dots \approx 103$ GBP.

Simple interest: $I = C \cdot r \cdot t$ with $C = 102.83$ GBP, $r = 0.06$ and $t = 0.5$ years (six months).
$I = 102.83 \cdot 0.06 \cdot 0.5 = 3.08 \dots$ GBP. If we start from the already rounded 103 GBP, we get $3.09$ .

Answer: a) (i) 491.5 euros; (ii) $\approx$ 103 GBP; b) about 3.08 GBP of interest (3.09 using 103 GBP).

Exercise 85 MediumIB · P1 TZ0 November 2003

David invests 6,000 Australian dollars (AUD) in a bank that offers 6% compound interest per year.

a) Calculate the amount of money he will have in 10 years.

b) After 10 years, David withdraws 5,000 AUD to invest in another bank that offers 8% compound interest per year. Calculate the total amount of money he will have in the two banks after one more year. Give your answer to the nearest Australian dollar.

View solution

Annual compound interest: $V = 6000 \cdot 1.0 6^{10}$ .
$V = 10745.09 \dots$

Answer a): approximately 10,745 AUD.

After withdrawing 5000 AUD, $10745.09 - 5000 = 5745.09$ AUD remain in the first bank, which after one more year at $6 %$ becomes $5745.09 \cdot 1.06 = 6089.80$ .
The 5000 AUD in the second bank, after one year at $8 %$ , becomes $5000 \cdot 1.08 = 5400$ .
Total: $6089.80 + 5400 = 11489.80 \approx 11490$ .

Answer b): 11,490 AUD.

Exercise 86 MediumIB · P1 TZ0 May 2000

Hassan invested 10,000 CHF in 1971. The interest rate was 5% per year. How much interest in total would Hassan have earned by the end of 1999 if

a) he had withdrawn the interest from his account at the end of each year;

b) he had not withdrawn the interest from his account at the end of each year.

Show answer

Answer: a) 14,000 CHF; b) $\approx 29, 200$ CHF.

Show working and answer

From 1971 to the end of 1999 there are 28 years.

a) If the interest is withdrawn each year, the capital remains 10,000 CHF and the same interest is earned each year.

Annual interest: $10000 \cdot 0.05 = 500$ CHF.
Over 28 years: $28 \cdot 500 = 14000$ CHF.

b) Without withdrawing the interest, compound interest applies.

Final value: $10000 \cdot 1.0 5^{28} \approx 39201.29$ CHF.
Total interest: $39201.29 - 10000 \approx 29201.29$ CHF, that is, about 29,200 CHF.

Answer: a) 14,000 CHF; b) $\approx 29, 200$ CHF.

Exercise 87 MediumIB · P2 TZ0 May 2000

Miranti deposits $ 1,000 into an investment account that pays 5% interest per year.

a) What will be the value of the investment after 5 years if the interest is reinvested?

b) How many years would it take for Miranti's $ 1,000 investment to double in value?

At the start of each year, Brenda deposits $ 1,000 into an investment account that pays 5% interest per year. The interest is calculated annually and reinvested.

c) How much will be in Brenda's account after 5 years?

Show answer

Answer: a) $ 1,276.28; b) 15 years; c) $ 5,801.91.

Show working and answer

a) With reinvestment the interest is compound: $V = 1000 \cdot 1.0 5^{5} = 1276.28$ $.

b) We look for $n$ with $1000 \cdot 1.0 5^{n} = 2000$ , that is, $1.0 5^{n} = 2$ .

We solve with the GDC (or logarithms): $n = \frac{ln 2}{ln 1.05} \approx 14.2$ .
As interest is paid for complete years, the investment exceeds double after 15 years (after 14 years it is still $1000 \cdot 1.0 5^{14} \approx 1979.93 < 2000$ ).

c) Each deposit of $ 1,000 is made at the start of a year, so the first one earns 5 years of interest, the second 4, and so on, down to the last one, which earns 1: $1000 \cdot 1.0 5^{5} + 1000 \cdot 1.0 5^{4} + 1000 \cdot 1.0 5^{3} + 1000 \cdot 1.0 5^{2} + 1000 \cdot 1.05.$

It is a geometric sequence with first term $1050$ , common ratio $1.05$ and 5 terms.
$S_{5} = \frac{1050 ( 1.0 5 ^{5} - 1 )}{1.05 - 1} \approx 5801.91$ $.

Answer: a) $ 1,276.28; b) 15 years; c) $ 5,801.91.

Exercise 88 MediumIB · P1 TZ0 November 2000

John invests $X$ USD in a bank. The interest rate set by the bank is 6% per year, compounded monthly.

a) Write down, in terms of $X$ , an expression for the value of John's investment after one year.

b) What interest rate, when compounded annually (instead of monthly), will give the same value of John's investment as in part (a)? Give your answer correct to three significant figures.

Show answer

Answer: a) $X \cdot 1.00 5^{12}$ ; b) $r \approx 6.17%$ .

Show working and answer

a) The interest is 6% per year compounded monthly: $6%/12 = 0.5%$ per month, with 12 periods per year. $V = X (1 + \frac{0.06}{12})^{12} = X \cdot 1.00 5^{12} .$

b) Let $r$ be the annual rate with annual compounding. It must hold that $X (1 + \frac{r}{100}) = X \cdot 1.00 5^{12} .$

We cancel $X$ : $1 + \frac{r}{100} = 1.00 5^{12} \approx 1.0616778$ .
Then $\frac{r}{100} \approx 0.0616778$ , so $r \approx 6.17%$ .

Answer: a) $X \cdot 1.00 5^{12}$ ; b) $r \approx 6.17%$ .

Exercise 89 HardIB · P2 TZ1 May 2018

Give your answers to parts (b), (c) and (d) correct to the nearest whole number.

Harinder has 14,000 US dollars (USD) to invest for a period of five years. He has two options for how to invest the money.

Option A: He invests the full amount, in USD, in a fixed deposit account at a US bank. The account pays a nominal annual interest rate of $r %$ , compounded annually, for five years. The bank manager says that this will give Harinder a return of 17,500 USD.

a) Calculate the value of $r$ .

Option B: He invests the full amount, in Indian rupees (INR), in a fixed deposit account at an Indian bank. The money must be converted from USD to INR before it is invested. The exchange rate is $1 USD = 66.91 INR$ .

b) Calculate 14,000 USD in INR.

The account at the Indian bank pays a nominal annual interest rate of $5.2%$ compounded monthly.

c) Calculate the amount of this investment, in INR, in this account after five years.

Harinder chose option B. After five years, Harinder converted this investment back into US DOLLARS. The exchange rate at that time was $1 USD = 67.16 INR$ .

d) Calculate how much more money, in USD, Harinder earned by choosing option B instead of option A.

View solution

Annual compound interest over $5$ years: $14000 (1 + \frac{r}{100})^{5} = 17500$ .
$(1 + \frac{r}{100})^{5} = \frac{17500}{14000} = 1.25$ .
$1 + \frac{r}{100} = 1.2 5^{1/5} = 1.04564$ , so $r = 4.56$ (this can also be obtained with the GDC's financial solver).

Answer a): $r \approx 4.56 %$ .

$14000 \cdot 66.91 = 936740$ .

Answer b): $936740$ INR.

Monthly interest: $n = 12$ , $t = 5$ years, $60$ periods, rate $5.2 %$ .
$F V = 936740 (1 + \frac{5.2}{1200})^{60} = 936740 \cdot 1.296202 \dots = 1214204$ .

Answer c): $1214204$ INR.

Converting back to USD: $\frac{1214204}{67.16} = 18 079.27 \dots \approx 18079$ USD.
Option A gave $17500$ USD.
Difference: $18079 - 17500 = 579$ .

Answer d): with option B, Harinder earned $579$ USD more.

Exercise 90 HardIB · P2 TZ0 November 2013

In this question, give all answers correct to two decimal places.

Arthur lives in London. On 1 August 2008 Arthur paid 37,500 euros (EUR) for a new car imported from Germany. The price of the same car in London was 34,075 pounds sterling (GBP). The exchange rate on 1 August 2008 was: 1 EUR $=$ 0.7234 GBP.

a) Calculate, in GBP, the price Arthur paid for the car.

b) Write down, in GBP, how much money Arthur saved by buying the car in Germany.

On 1 August 2008 Arthur invested the money he had saved in a bank that paid simple annual interest of 4.5%.

c) Calculate, in GBP, the value of Arthur's investment on 1 August 2012.

Between 1 August 2008 and 1 August 2012 Arthur's car depreciated at an annual rate of 9% of its value.

d) Calculate, in GBP, the value of Arthur's car on 1 August 2009.

e) Verify that the value of Arthur's car on 1 August 2012 was 18,600 GBP, rounded to the nearest 100 GBP.

On 1 August 2012 Arthur sold his car for 18,600 GBP and bought a new car imported from Germany, for which he paid 30,500 EUR. To buy the new car, he used the 18,600 GBP and the value of the investment he had made on 1 August 2008.

The exchange rate on 1 August 2012 was: 1 EUR $=$ 0.8694 GBP.

f) Calculate how much money, in EUR, is left over after buying the car.

Show answer

Answer: a) $27127.50$ GBP; b) $6947.50$ GBP; c) $8198.05$ GBP; d) $24686.03$ GBP; e) verified ( $\approx 18600$ GBP); f) $323.61$ EUR.

Show working and answer

a) We convert euros to pounds: $1 EUR = 0.7234$ GBP.

$37500 \cdot 0.7234 = 27127.50$ .

Answer a): $27127.50$ GBP.

b) The saving is the difference between the London price and what he paid:

$34075 - 27127.50 = 6947.50$ .

Answer b): $6947.50$ GBP.

c) Simple interest of $4.5%$ per year for 4 years (from August 2008 to August 2012).

Annual interest: $6947.50 \cdot 0.045 = 312.6375$ .
In 4 years: $312.6375 \cdot 4 = 1250.55$ .
Value: $6947.50 + 1250.55 = 8198.05$ . (This is equivalent to $6947.50 \cdot 1.18$ .)

Answer c): $8198.05$ GBP.

d) The car cost $27127.50$ GBP and depreciates by $9%$ per year, that is, each year it is worth $91%$ of the previous year.

After one year: $27127.50 \cdot 0.91 = 24686.025$ .

Answer d): $24686.03$ GBP.

e) After 4 years (from 2008 to 2012):

$27127.50 \cdot 0.9 1^{4} = 27127.50 \cdot 0.685750 \approx 18602.67$ GBP.
Rounded to the nearest 100 GBP: $18600$ GBP.

Verified.

f) Money available on 1 August 2012, in GBP:

$18600 + 8198.05 = 26798.05$ GBP.
Price of the new car in GBP: $30500 \cdot 0.8694 = 26516.70$ GBP.
Left over: $26798.05 - 26516.70 = 281.35$ GBP.
We convert to euros by dividing by $0.8694$ : $\frac{281.35}{0.8694} = 323.61$ EUR.

Answer: a) $27127.50$ GBP; b) $6947.50$ GBP; c) $8198.05$ GBP; d) $24686.03$ GBP; e) verified ( $\approx 18600$ GBP); f) $323.61$ EUR.

Exercise 91 HardIB · P2 TZ1 May 2011

Give all answers to this question to the nearest whole unit of currency.

In January 2008, Larry had 90,000 dollars to invest for his retirement in January 2011. He invested 40,000 USD in US government bonds that paid 4% per annum simple interest.

a) Calculate the value of Larry's investment in government bonds in January 2011.

Larry exchanged this investment into South African rand (ZAR) at an exchange rate of 1 USD $=$ 18.624 ZAR.

b) Calculate the amount Larry received in ZAR from the exchange.

He exchanged the remaining 50,000 USD into South African rand (ZAR) in January 2008.

The exchange rate between USD and ZAR was 1 USD $=$ 10.608 ZAR. A commission of 2.5% was charged on the exchange.

c) Calculate the value, in USD, of the commission Larry paid.

d) Show that the amount Larry had to invest is 517,000 ZAR, correct to the nearest 1,000 ZAR.

In January 2008, Larry deposited this money in a bank account that paid interest at a nominal annual rate of 12%, compounded monthly.

e) Find the value of Larry's money in the bank account in January 2011.

Show answer

Answer: a) $44800$ USD; b) $834355$ ZAR; c) $1250$ USD; d) proved; e) $739707$ ZAR.

Show working and answer

a) Simple interest of $4%$ for 3 years (January 2008 to January 2011).

Annual interest: $40000 \cdot 0.04 = 1600$ .
Over 3 years: $4800$ .
Value: $40000 + 4800 = 44800$ .

Answer a): $44800$ USD.

b) Exchange at $1 USD = 18.624$ ZAR:

$44800 \cdot 18.624 = 834355.2$ .

Answer b): $834355$ ZAR.

c) The commission is $2.5%$ of the $50000$ USD exchanged:

$50000 \cdot 0.025 = 1250$ .

Answer c): $1250$ USD.

d) We exchange only what remains after paying the commission:

Money exchanged: $50000 - 1250 = 48750$ USD.
In rand: $48750 \cdot 10.608 = 517140$ ZAR.
To the nearest 1,000 ZAR: $517000$ ZAR.

Proved.

e) Compound monthly interest, $k = 12$ , $r = 12%$ , over 3 years: $36$ periods, monthly rate of $1%$ .

$F V = 517000 (1 + \frac{12}{1200})^{36} = 517000 \cdot 1.0 1^{36}$ .
Using the GDC: $F V \approx 739707$ .

Answer: a) $44800$ USD; b) $834355$ ZAR; c) $1250$ USD; d) proved; e) $739707$ ZAR.

Exercise 92 HardIB · P2 TZ2 May 2011

Give all your numerical answers correct to two decimal places.

On 1 January 2005, Daniel invested 30,000 AUD at a simple annual interest rate in a Regular Saver account. On 1 January 2007, Daniel had 31,650 AUD in the account.

a) Calculate the interest rate.

On 1 January 2005, Rebecca invested 30,000 AUD in a Supersaver account with a nominal annual rate of 2.5%, compounded annually.

b) Calculate the amount in the Supersaver account after two years.

c) Find the number of complete years from 1 January 2005 for the amount in Rebecca's account to exceed the amount in Daniel's account.

On 1 January 2007, Daniel reinvested 80% of the money in his Regular Saver account into an Extra Saver account at a nominal annual rate of 3%, compounded quarterly.

d) (i) Calculate the amount of money reinvested by Daniel on 1 January 2007.

(ii) Find the number of complete years needed for Daniel's amount in the Extra Saver account to exceed 30,000 AUD.

Show answer

Answer: a) $2.75%$ ; b) $31518.75$ AUD; c) $9$ years; d) (i) $25320$ AUD; (ii) $6$ complete years.

Show working and answer

a) Simple interest: over 2 years (2005 to 2007) $31650 - 30000 = 1650$ AUD were earned.

Annual interest: $\frac{1650}{2} = 825$ AUD.
Rate: $\frac{825}{30000} \cdot 100 = 2.75$ .

Answer a): $2.75%$ .

b) Compound annual interest of $2.5%$ , 2 years:

$30000 \cdot 1.02 5^{2} = 31518.75$ .

Answer b): $31518.75$ AUD.

c) We compare the two accounts year by year. Daniel: $30000 + 825 n$ . Rebecca: $30000 \cdot 1.02 5^{n}$ . We solve using the GDC (table of values or graphs of both functions).

$n = 8$ : Daniel $36600$ ; Rebecca $36552.08$ . Rebecca does not yet exceed it.
$n = 9$ : Daniel $37425$ ; Rebecca $37465.89$ . Rebecca now exceeds it.

Answer c): $9$ years.

d) (i) $80%$ of $31650$ is reinvested:

$0.80 \cdot 31650 = 25320$ .

(ii) Compound quarterly interest, $k = 4$ , $r = 3%$ : quarterly rate $0.75%$ . We want $25320 \cdot 1.007 5^{4 n} > 30000$ .

Using the GDC we obtain $4 n > 22.7$ , that is, 23 quarters are needed ( $5.75$ years).
In complete years: after $5$ years ( $20$ quarters) there are $\approx 29401$ AUD, which is not enough; after $6$ years ( $24$ quarters) there are $\approx 30293$ AUD, which exceeds $30000$ .

Answer: a) $2.75%$ ; b) $31518.75$ AUD; c) $9$ years; d) (i) $25320$ AUD; (ii) $6$ complete years.

Exercise 93 HardIB · P2 TZ2 May 2010

In this question, give all answers correct to the nearest whole unit of the relevant currency.

Ying and Ruby each have 5,000 USD to invest.

Ying invests 5,000 USD in a bank account that pays a nominal annual interest rate of 4.2%, compounded annually. Ruby invests 5,000 USD in an account that offers a fixed interest of 230 USD each year.

a) Find how much money Ruby will have in the bank after 3 years.

b) Verify that Ying will have 7,545 USD in the bank after 10 years.

c) Find the number of complete years that must pass for Ying's investment to exceed 6,500 USD for the first time.

d) Find the number of complete years that must pass for Ying's investment to exceed Ruby's investment.

Ruby moves to live in Italy from the United States. Therefore, she transfers 6,610 USD to an Italian bank which applies the following exchange rate: 1 USD $=$ 0.735 euros. The bank charges Ruby a commission of 1.8%.

e) Calculate how much money Ruby will have to invest at the Italian bank, after paying this commission.

Ruby returns to the United States on holiday. She exchanges 800 euros at a bank in Chicago and receives 1,006.20 USD. The bank has announced that it applies the following exchange rate: 1 euro $=$ 1.29 USD.

f) Calculate the commission, expressed as a percentage, charged to Ruby at this bank.

Show answer

Answer: a) 5690 USD; b) verified; c) 7 years; d) 6 years; e) 4771 €; f) 2.5%

Show working and answer

a) The interest is fixed: 230 USD each year.

Money after 3 years: $5000 + 3 \cdot 230 = 5690$ .

Answer a): 5690 USD

b) Compound annual interest: $F V = P V (1 + \frac{r}{100 k})^{k n}$ with $k = 1$ .

$F V = 5000 \cdot (1.042)^{10}$
$F V = 5000 \cdot 1.50896 \dots = 7544.8 \approx 7545$ USD.

Verified.

c) We look for the smallest integer $n$ with $5000 \cdot 1.04 2^{n} > 6500$ .

We divide by 5000: $1.04 2^{n} > 1.3$ .
Using the GDC (table or solving the equation) $1.04 2^{n} = 1.3$ gives $n \approx 6.38$ .
Check: with $n = 6$ we get 6,400 USD (does not exceed); with $n = 7$ we get 6,668 USD (does exceed).

Answer c): 7 years

d) Ruby's value after $n$ years: $5000 + 230 n$ . Ying's value: $5000 \cdot 1.04 2^{n}$ . We compare using the GDC.

$n = 5$ : Ying $= 6142$ ; Ruby $= 6150$ . Ying does not yet exceed Ruby.
$n = 6$ : Ying $= 6400$ ; Ruby $= 6380$ . Ying now exceeds Ruby.

Answer d): 6 years

e) 1. We convert to euros: $6610 \cdot 0.735 = 4858.35$ €. 2. The commission is 1.8%, so 98.2% remains: $4858.35 \cdot 0.982 = 4770.90$ .

Answer e): 4771 €

f) 1. Without commission she would receive $800 \cdot 1.29 = 1032$ USD. 2. Ruby receives 1,006.20 USD, so the commission is $1032 - 1006.20 = 25.80$ USD. 3. Percentage: $\frac{25.80}{1032} \cdot 100 = 2.5$ .

Answer: a) 5690 USD; b) verified; c) 7 years; d) 6 years; e) 4771 €; f) 2.5%

Exercise 94 HardIB · P2 TZ0 November 2010

In this question, give all answers correct to two decimal places.

Section A

Estela lives in Brazil and wants to exchange 4,000 BRL (Brazilian reais) into GBP (British pounds sterling). The exchange rate is 1.00 BRL $=$ 0.3071 GBP. The bank charges a 3% commission on the amount in BRL.

a) Find, in BRL, how much money Estela has after paying the commission.

b) Find, in GBP, how much money Estela receives.

After her trip to the United Kingdom, Estela has 400 GBP left over. At the airport she exchanges this money back into BRL. The exchange rate is now 1.00 BRL $=$ 0.3125 GBP.

c) Find, in BRL, how much money Estela should receive.

In reality, and after paying a commission, Estela ends up receiving 1,216.80 BRL.

d) Find, in BRL, the commission charged to Estela.

e) The commission percentage is $t %$ . Find the value of $t$ .

Section B

Daniel invests $1,000 (US dollars) in an account that offers a nominal annual interest rate of 3.5%, compounded semi-annually.

a) Show that after three years Daniel will have $1,109.70 in his account (rounded to two decimal places).

b) Write down the interest Daniel receives after these three years.

Helen invests $1,000 in an account that offers simple annual interest.

c) Find the annual simple interest rate required for Helen to have $1,109.70 in her account after three years.

Show answer

Answer: A) a) 3880.00 BRL; b) 1191.55 GBP; c) 1280.00 BRL; d) 63.20 BRL; e) $t = 4.94$ . B) a) verified; b) $109.70; c) $r = 3.66 %$

Show working and answer

Section A

a) 1. Commission: 3% of 4000 $= 120$ BRL. 2. Remaining: $4000 - 120 = 3880.00$ BRL.

Answer a): 3880.00 BRL

b) 1. $3880 \cdot 0.3071 = 1191.548$

Answer b): 1191.55 GBP

c) 1. As 1 BRL $= 0.3125$ GBP, we divide: $\frac{400}{0.3125} = 1280$ .

Answer c): 1280.00 BRL

d) 1. Commission $= 1280 - 1216.80 = 63.20$ .

Answer d): 63.20 BRL

e) 1. $t = \frac{63.20}{1280} \cdot 100 = 4.9375$

Answer e): $t = 4.94$

Section B

a) Semi-annual compounding: $k = 2$ , $n = 3$ , rate per period $\frac{3.5}{2} = 1.75 %$ .

$F V = 1000 (1 + \frac{3.5}{200})^{6} = 1000 \cdot 1.017 5^{6}$
$F V = 1000 \cdot 1.10970 \dots = 1109.70$

Verified.

b) 1. Interest $= 1109.70 - 1000 = 109.70$ .

Answer b): $109.70

c) Simple interest: $I = P V \cdot \frac{r}{100} \cdot n$ .

$109.70 = 1000 \cdot \frac{r}{100} \cdot 3$
$r = \frac{109.70}{30} = 3.6566 \dots$

Answer: A) a) 3880.00 BRL; b) 1191.55 GBP; c) 1280.00 BRL; d) 63.20 BRL; e) $t = 4.94$ . B) a) verified; b) $109.70; c) $r = 3.66 %$

Exercise 95 HardIB · P2 TZ0 November 2001

The following table shows the deposits, in Australian dollars (AUD), made by Vicki into an investment account on the first day of each month for the first four months of 1999. The interest rate is 0.75% per month, compounded monthly. Interest is added to the account at the end of each month.

Month	Deposit (AUD)
January	600
February	1,300
March	230
April	710

a) Show that the amount of money in Vicki's account at the end of February is AUD 1,918.78.

b) Calculate the amount of Australian dollars in Vicki's account at the end of April.

Vicki makes no withdrawals or deposits after 1 April 1999.

c) How much money is in Vicki's account at the end of December 1999?

From 1 January 2000, the bank applies a new interest rate of 3.5% per year, compounded annually.

d) After how many complete years after December 1999 will Vicki's investment exceed AUD 3,300?

View solution

On 1 January she deposits $600$ ; at the end of January, with monthly interest of $0.75 %$ : $600 \cdot 1.0075 = 604.50$ .
On 1 February she adds $1300$ : $604.50 + 1300 = 1904.50$ .
At the end of February: $1904.50 \cdot 1.0075 = 1918.78 \dots$

Shown.

On 1 March: $1918.78 + 230 = 2148.78$ . End of March: $2148.78 \cdot 1.0075 = 2164.90$ (without rounding intermediate values: $2164.898 \dots$ ).
On 1 April: $2164.898 + 710 = 2874.898$ . End of April: $2874.898 \cdot 1.0075 = 2896.46 \dots$

Answer b): AUD 2,896.46.

From the end of April to the end of December there are $8$ months with no transactions: $2896.46 \cdot 1.007 5^{8}$ .
$= 3074.88 \dots$

Answer c): AUD 3,074.88.

From January 2000, annual interest of $3.5 %$ : after $n$ years, $3074.88 \cdot 1.03 5^{n}$ .
We look for the smallest whole number $n$ with $3074.88 \cdot 1.03 5^{n} > 3300$ . Using the GDC (table of values or solving the equation), equality occurs at $n \approx 2.05$ .
Check: $n = 2$ gives $3074.88 \cdot 1.03 5^{2} = 3293.9 < 3300$ ; $n = 3$ gives $3074.88 \cdot 1.03 5^{3} = 3409.3 > 3300$ .

Answer d): 3 complete years.

Exercise 96 HardIB · P2 TZ0 November 2005

Ali, Bob and Connie each want to invest 3,000 USD (US dollars).

Ali invests the 3,000 USD in an institution that offers simple interest at a rate of 4.5% per year. The interest is paid at the end of each year.

Bob invests the 3,000 USD in a bank that offers interest compounded annually at a rate of 4% per year. The interest is paid at the end of each year.

Connie invests the 3,000 USD in another bank that offers interest compounded semi-annually at a rate of 3.8% per year. The interest is paid at the end of each half-year.

a) Calculate the amount of money that Ali and Bob will have at the start of the seventh year.

b) Verify that, at the start of the seventh year, Connie has 3,760.20 USD.

c) Calculate how many years it will take Bob to have 6,000 USD in the bank.

At the start of the seventh year, Connie moves to England. Connie makes a transfer of her money to a bank there, at an exchange rate of 1 USD = 0.711 GBP (British pounds sterling). The bank charges a 2% commission.

d) (i) Calculate, in USD, the commission charged by the bank.

(ii) Calculate the amount of money, in GBP, that Connie transfers to the bank in England.

Show answer

Answer: a) Ali 3,810 USD, Bob $\approx$ 3,796 USD; b) verified; c) $\approx 17.7$ years; d) (i) 75.20 USD; (ii) $\approx 2, 620.04$ GBP.

Show working and answer

a) At the start of the seventh year, 6 complete years have passed.

Ali (simple interest): each year Ali receives $3000 \cdot 0.045 = 135$ USD. After 6 years: $3000 + 6 \cdot 135 = 3810$ USD.
Bob (annual compound interest): $V = 3000 \cdot (1 + \frac{4}{100})^{6} = 3000 \cdot 1.0 4^{6} \approx 3795.96$ USD, that is, about 3796 USD.

b) Connie compounds every half-year, so in 6 years there are $6 \cdot 2 = 12$ periods, with a rate of $3.8%/2 = 1.9%$ per half-year.

$V = 3000 \cdot (1 + \frac{3.8}{200})^{12} = 3000 \cdot 1.01 9^{12}$ .
Using the calculator: $3000 \cdot 1.01 9^{12} \approx 3760.20$ USD.

Verified.

c) We look for $n$ such that $3000 \cdot 1.0 4^{n} = 6000$ , that is, $1.0 4^{n} = 2$ .

We solve the equation with the GDC's solver (or with logarithms): $n = \frac{ln 2}{ln 1.04} \approx 17.7$ .
About 17.7 years are needed (in practice, the value exceeds 6,000 USD during year 18).

d) (i) Commission: $2%$ of $3760.20$ , that is, $0.02 \cdot 3760.20 = 75.20$ USD.

(ii) 1. Money left after the commission: $3760.20 - 75.20 = 3685.00$ USD. 2. Conversion to pounds: $3685 \cdot 0.711 = 2620.035 \approx 2620.04$ GBP.

Answer: a) Ali 3,810 USD, Bob $\approx$ 3,796 USD; b) verified; c) $\approx 17.7$ years; d) (i) 75.20 USD; (ii) $\approx 2, 620.04$ GBP.

Depreciation and growth

Exercise 97 MediumIB · P1 TZ0 November 2015

The population of the city of Nouméa at the end of 2008 was 97,579 inhabitants. One year later, at the end of 2009, the population of Nouméa was 99,921 inhabitants.

a) Find the percentage change in the population from the end of 2008 to the end of 2009.

Now suppose that the population of Nouméa continues to grow at the same percentage rate of change.

b) Use your answer to part (a) to estimate the population at the end of 2016.

c) Write your answer to part (b) in the form $a \times 1 0^{k}$ , where $1 \leq a < 10, k \in Z$ .

Show answer

Answer: a) $2.40%$ ; b) $\approx 117966$ inhabitants; c) $1.18 \times 1 0^{5}$ .

Show working and answer

a) Percentage change = $\frac{final value - initial value}{initial value} \cdot 100$ .

$\frac{99921 - 97579}{97579} \cdot 100 = \frac{2342}{97579} \cdot 100 \approx 2.40$ .

Answer a): an increase of $2.40%$ .

b) From the end of 2009 to the end of 2016 is $7$ years, with an annual growth of $2.4%$ (factor $1.024$ ).

Population in 2016: $99921 \cdot (1.024)^{7}$ .
$1.02 4^{7} \approx 1.18059$ , so the population is $\approx 117966$ inhabitants.

Answer b): approximately $117966$ inhabitants.

c) Standard form with $1 \leq a < 10$ :

$117966 = 1.17966 \times 1 0^{5} \approx 1.18 \times 1 0^{5}$ .

Answer: a) $2.40%$ ; b) $\approx 117966$ inhabitants; c) $1.18 \times 1 0^{5}$ .

Exercise 98 MediumIB · P1 TZ1 May 2008

In an experiment it was found that a bacterial culture triples in number every four hours. There are 200 bacteria at the start of the experiment.

Hours	0	4	8	12	16
Number of bacteria	200	600	$a$	5400	16200

a) Find the value of $a$ .

b) Calculate how many bacteria there will be after one day.

c) Find how long it will take for there to be two million bacteria.

Show answer

Answer: a) 1800; b) 145,800; c) about 33.5 hours

Show working and answer

a) Every 4 hours it is multiplied by 3: $a = 600 \cdot 3 = 1800$ . (Check: $1800 \cdot 3 = 5400$ .)

Answer a): $a = 1800$

b) One day is 24 hours, that is, $24/4 = 6$ periods.

$N = 200 \cdot 3^{6} = 200 \cdot 729 = 145800$

Answer b): 145,800 bacteria

c) With $t$ hours there are $N = 200 \cdot 3^{t /4}$ bacteria.

$200 \cdot 3^{t /4} = 2000000 \Rightarrow 3^{t /4} = 10000$
Taking logarithms (or solving with the GDC): $\frac{t}{4} = \frac{ln 10000}{ln 3} = 8.384$
$t = 4 \cdot 8.384 = 33.5$

Answer: a) 1800; b) 145,800; c) about 33.5 hours

Exercise 99 MediumIB · P1 TZ0 May 2003

Mario has spent $ 40,000 on buying some land. The land increases in value by 5% every year.

(i) What is the value of the land after five years?

After five years, Mario sells the land. He pays a 1% tax on the sale and spends the rest of the money on a car. The car depreciates at a rate of $ 2,500 per year.

(ii) How much tax does Mario pay?

(iii) What is the value of the car five years after Mario bought it?

Show answer

Answer: (i) $ 51,051.26; (ii) $ 510.51; (iii) $ 38,040.75.

Show working and answer

(i)

Annual increase of 5%: a factor of $1.05$ each year, so $V = 40000 \cdot 1.0 5^{5}$ .
$V = 51051.26 \dots$

(ii)

The tax is $1 %$ of the sale: $0.01 \cdot 51051.26 = 510.51$ .

(iii)

Money left for the car: $51051.26 - 510.51 = 50540.75$ (or $0.99 \cdot 51051.26$ ).
It loses $ 2,500 per year (arithmetic sequence), so in 5 years it loses $5 \cdot 2500 = 12500$ .
Value: $50540.75 - 12500 = 38040.75$ .

Answer: (i) $ 51,051.26; (ii) $ 510.51; (iii) $ 38,040.75.

Exercise 100 HardIB · P2 TZ0 November 2008

Throughout this question, give all answers correct to the nearest whole number.

Parque School started classes in January 2000 with 100 students. Each complete year, the number of students increases by 6%.

a) Find the number of students at Parque School in

(i) January 2001;

(ii) January 2003.

b) Verify that in January 2007 the number of students at Parque School is equal to 150.

Arboleda School had 110 students in January 2000. Each complete year, the number of students increases by 10 compared to the previous year.

c) Find the number of students at Arboleda School in January 2003.

d) Find the first year in which the number of students at Arboleda School will be 60% more than in January 2000.

Each January, one of the two schools, the one with more students, is granted extra money to spend on sports equipment.

e) (i) Decide which school receives this money in 2007. Justify your answer.

(ii) Find the year in which Parque School will receive this extra money for the first time.

View solution

The number of students at Parque School grows by 6% each year: it is a geometric sequence with first term $100$ and common ratio $r = 1.06$ , that is, $P_{n} = 100 \cdot 1.0 6^{n}$ , where $n$ is the number of years since January 2000. Arboleda School grows by 10 students each year: it is an arithmetic sequence, $A_{n} = 110 + 10 n$ .

(i) From January 2000 to January 2001 is $n = 1$ year: $P_{1} = 100 \cdot 1.06 = 106$ .
(ii) From 2000 to 2003 is $n = 3$ years: $P_{3} = 100 \cdot 1.0 6^{3} = 119.1016 \approx 119$ .

Answer a): (i) 106 students; (ii) 119 students.

From 2000 to 2007 is $n = 7$ years.
$P_{7} = 100 \cdot 1.0 6^{7} = 150.363 \dots$
Rounding to the nearest whole number, $P_{7} = 150$ .

Verified.

From 2000 to 2003, $n = 3$ .
$A_{3} = 110 + 10 \cdot 3 = 140$ .

Answer c): 140 students.

60% more than January 2000 means $110 \cdot 1.6 = 176$ students.
We set $110 + 10 n \geq 176$ , so $10 n \geq 66$ and $n \geq 6.6$ .
As $n$ is a whole number of years, the smallest value is $n = 7$ (for $n = 6$ there are only 170 students, which does not reach 176; for $n = 7$ there are 180).
$n = 7$ corresponds to January $2000 + 7 = 2007$ .

Answer d): the year 2007.

(i) In 2007: Parque has $150$ students and Arboleda has $110 + 10 \cdot 7 = 180$ . As $180 > 150$ , the money goes to Arboleda School.
(ii) We look for the first $n$ with $100 \cdot 1.0 6^{n} > 110 + 10 n$ . We use the GDC (table of values for both functions, or intersection of their graphs), or we test values:
- $n = 18$ : $P_{18} = 285.4 \dots$ and $A_{18} = 290$ . Arboleda still leads.
- $n = 19$ : $P_{19} = 302.56 \dots$ and $A_{19} = 300$ . Parque now leads.
The first $n$ at which Parque exceeds Arboleda is $n = 19$ , which corresponds to January $2000 + 19 = 2019$ .

Answer e): (i) Arboleda School; (ii) Parque School receives this money for the first time in the year 2019.

Currency conversion

Exercise 101 BasicIB · P1 TZ0 November 2019

In this question, give all answers correct to the nearest whole number. Criselda travelled to Kota Kinabalu, in Malaysia. At the airport she saw the following information at the currency exchange office.

Currency exchange – Malaysian ringgit (MYR)	Sell	Buy
1 Singapore dollar (SGD)	3.22	3.07
1 US dollar (USD)	4.45	4.25

This means that at this currency exchange office they buy USD from travellers and, in return, give them MYR at the exchange rate 1 USD = 4.25 MYR. No commission is charged. Criselda exchanged 460 SGD into MYR.

a) Calculate how many MYR Criselda received.

During her time in Kota Kinabalu, Criselda spent 440 MYR. She returned to the currency exchange office and exchanged the rest of her MYR for USD.

b) Calculate how many USD Criselda received.

View solution

The office buys SGD from Criselda and gives her MYR at the "Buy" rate: $1$ SGD $= 3.07$ MYR.
$460 \cdot 3.07 = 1412.2$ , which rounded to the nearest whole number is $1412$ .

Answer a): $1412$ MYR.

She has $1412 - 440 = 972$ MYR left.
Now the office sells USD to Criselda and she hands over MYR: the "Sell" rate is used, $1$ USD $= 4.45$ MYR.
$\frac{972}{4.45} = 218.4 \dots \approx 218$ .

Answer b): $218$ USD.

$n$	$C_{n}$	$T_{n}$
15	1,950	1,906.35
16	2,160	2,156.98
17	2,380	2,432.68

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IB exam questions for topic 2

General term and properties of sequences

Arithmetic sequences: term and sum

Geometric sequences: term and sum

Simple and compound interest

Depreciation and growth

Currency conversion

Other sequence problems

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