IB Analysis and Approaches HL · Arithmetic and geometric sequences · End of topic

IB exam questions: topic 3

85 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.

Level

Paper

Solutions

Sequences and sigma notation

Exercise 1 BasicIB · P2 TZ2 May 2010

Consider the arithmetic sequence $8, 26, 44, \dots$

a) Find an expression for the $n$ th term.

b) Write down the sum of the first $n$ terms, using sigma notation.

c) Calculate the sum of the first 15 terms.

View solution

a) The common difference is $d = 26 - 8 = 18$ . Then $u_{n} = u_{1} + (n - 1) d = 8 + 18 (n - 1) = 18 n - 10.$

Answer a): $u_{n} = 18 n - 10$

b) The sum of the first $n$ terms is the sum of $u_{k}$ for $k = 1, \dots, n$ : $S_{n} = k = 1 \sum n (18 k - 10) .$ If we want to simplify: $S_{n} = \frac{n}{2} (8 + 18 n - 10) = 9 n^{2} - n$ .

Answer b): $S_{n} = k = 1 \sum n (18 k - 10) = 9 n^{2} - n$

c) With $n = 15$ : $u_{15} = 18 \cdot 15 - 10 = 260$ . Then $S_{15} = \frac{15}{2} (u_{1} + u_{15}) = \frac{15}{2} (8 + 260) = \frac{15 \cdot 268}{2} = 2010.$

Answer c): $S_{15} = 2010$

Exercise 2 BasicIB · P1 May 2000

The sum of the first $n$ terms of an arithmetic sequence is $S_{n} = 3 n^{2} - 2 n$ . Calculate the $n$ th term, $u_{n}$ .

Show answer

Answer: $u_{n} = 6 n - 5$

Show working and answer

For $n = 1$ : $u_{1} = S_{1} = 3 - 2 = 1$ .
For $n \geq 2$ we use $u_{n} = S_{n} - S_{n - 1}$ : $u_{n} = 3 n^{2} - 2 n - [3 (n - 1)^{2} - 2 (n - 1)] = 3 n^{2} - 2 n - 3 n^{2} + 6 n - 3 + 2 n - 2 = 6 n - 5.$
For $n = 1$ the formula gives $6 - 5 = 1$ , which agrees with $u_{1}$ . It is valid for all $n$ (arithmetic sequence with $u_{1} = 1$ and $d = 6$ ).

Answer: $u_{n} = 6 n - 5$

Exercise 3 BasicIB · P1 November 2002

Find $r = 1 \sum 50 ln (2^{r})$ , giving the answer in the form $a ln 2$ , where $a \in Q$ .

Show answer

Answer: $1275 ln 2$ (that is, $a = 1275$ )

Show working and answer

We apply the logarithm of a power property, $ln (2^{r}) = r ln 2$ .
Taking the constant factor out of the sum: $r = 1 \sum 50 ln (2^{r}) = ln 2 r = 1 \sum 50 r$ .
The sum of the first $n$ integers is $\frac{n ( n + 1 )}{2}$ , so $r = 1 \sum 50 r = \frac{50 \cdot 51}{2} = 1275$ .
Therefore the sum equals $1275 ln 2$ .

Answer: $1275 ln 2$ (that is, $a = 1275$ )

Exercise 4 MediumIB · P2 TZ2 May 2013

(i) Express the sum of the first $n$ positive odd integers using sigma notation.

(ii) Verify that the sum in part (i) is equal to $n^{2}$ .

(iii) Deduce the value of the difference between the sum of the first 47 positive odd integers and the sum of the first 14 positive odd integers.

Show answer

Answer: $2013$

Show working and answer

(i)

The $k$ th positive odd integer is $2 k - 1$ (1, 3, 5, ...). Therefore the sum is $k = 1 \sum n (2 k - 1) .$

(ii)

It is the sum of an arithmetic sequence with $u_{1} = 1$ and $d = 2$ .
$S_{n} = \frac{n}{2} [2 \cdot 1 + (n - 1) \cdot 2] = \frac{n}{2} \cdot 2 n = n^{2}$ .

Hence proved.

(iii)

The sum of the first 47 is $4 7^{2} = 2209$ and the sum of the first 14 is $1 4^{2} = 196$ .
Difference: $2209 - 196 = 2013$ .

Answer: $2013$

Exercise 5 MediumIB · P2 November 2002

A sequence ${u_{n}}$ is given by $u_{0} = 1$ , $u_{1} = 2$ , $u_{n + 1} = 3 u_{n} - 2 u_{n - 1}$ where $n \in Z^{+}$ .

a) Find $u_{2}$ , $u_{3}$ , $u_{4}$ .

b) (i) Express $u_{n}$ in terms of $n$ .

(ii) Verify that your answer to (b)(i) satisfies the equation $u_{n + 1} = 3 u_{n} - 2 u_{n - 1}$ .

Show answer

Answer: a) $u_{2} = 4$ , $u_{3} = 8$ , $u_{4} = 16$ ; b) $u_{n} = 2^{n}$

Show working and answer

a) 1. With $n = 1$ : $u_{2} = 3 u_{1} - 2 u_{0} = 3 \cdot 2 - 2 \cdot 1 = 4$ .

With $n = 2$ : $u_{3} = 3 u_{2} - 2 u_{1} = 12 - 4 = 8$ .
With $n = 3$ : $u_{4} = 3 u_{3} - 2 u_{2} = 24 - 8 = 16$ .

b) (i) 1. The sequence $1, 2, 4, 8, 16$ suggests $u_{n} = 2^{n}$ , which satisfies $u_{0} = 1$ and $u_{1} = 2$ .

Justification: the characteristic equation $r^{2} = 3 r - 2$ , that is, $r^{2} - 3 r + 2 = 0$ , has roots $1$ and $2$ ; thus $u_{n} = A + B \cdot 2^{n}$ , and with $u_{0} = 1$ , $u_{1} = 2$ we get $A + B = 1$ , $A + 2 B = 2$ , so $B = 1$ , $A = 0$ .

(ii) 1. Substituting $u_{n} = 2^{n}$ into the right-hand side: $3 \cdot 2^{n} - 2 \cdot 2^{n - 1} = 3 \cdot 2^{n} - 2^{n} = 2 \cdot 2^{n} = 2^{n + 1}$ .

This is $u_{n + 1}$ , so the equation holds.

Answer: a) $u_{2} = 4$ , $u_{3} = 8$ , $u_{4} = 16$ ; b) $u_{n} = 2^{n}$

Exercise 6 MediumIB · P1 TZ0 November 2004

The sum of the first $n$ terms of a series is given by $S_{n} = 2 n^{2} - n$ , $n \in Z^{+}$ .

a) Find the first three terms of the series.

b) Find an expression for the $n$ th term of the series, giving your answer in terms of $n$ .

Show answer

Answer: a) $1, 5, 9$ ; b) $u_{n} = 4 n - 3$

Show working and answer

a) 1. The first term is $u_{1} = S_{1} = 2 (1)^{2} - 1 = 1$ . 2. $S_{2} = 2 (2)^{2} - 2 = 6$ , so $u_{2} = S_{2} - S_{1} = 6 - 1 = 5$ . 3. $S_{3} = 2 (3)^{2} - 3 = 15$ , so $u_{3} = S_{3} - S_{2} = 15 - 6 = 9$ .

The first three terms are $1, 5, 9$ .

b) 1. For $n \geq 2$ , the $n$ th term is $u_{n} = S_{n} - S_{n - 1}$ . 2. We calculate $S_{n - 1} = 2 (n - 1)^{2} - (n - 1) = 2 n^{2} - 4 n + 2 - n + 1 = 2 n^{2} - 5 n + 3$ . 3. Then $u_{n} = (2 n^{2} - n) - (2 n^{2} - 5 n + 3) = 4 n - 3$ . 4. For $n = 1$ the formula gives $4 - 3 = 1 = u_{1}$ , so it is valid for all $n \in Z^{+}$ . (It agrees with $1, 5, 9$ : an arithmetic sequence with common difference $4$ .)

Answer: a) $1, 5, 9$ ; b) $u_{n} = 4 n - 3$

Exercise 7 MediumIB · P1 TZ0 November 2005

Find $n = 1 \sum 15 a_{n}^{2}$ , where $a_{n} = ln x^{n}$ .

Show answer

Answer: $1240 (ln x)^{2}$

Show working and answer

By the properties of logarithms, $a_{n} = ln x^{n} = n ln x$ .
Then $a_{n}^{2} = n^{2} (ln x)^{2}$ and $n = 1 \sum 15 a_{n}^{2} = (ln x)^{2} n = 1 \sum 15 n^{2}$ .
Using the formula $n = 1 \sum N n^{2} = \frac{N ( N + 1 ) ( 2 N + 1 )}{6}$ with $N = 15$ : $\frac{15 \cdot 16 \cdot 31}{6} = 1240$ .
Hence the sum is $1240 (ln x)^{2}$ .

Answer: $1240 (ln x)^{2}$

Exercise 8 HardIB · P2 TZ0 November 2005Proof

a) Prove by mathematical induction that $r = 1 \sum n \frac{1}{( 2 r - 1 ) ( 2 r + 1 )} = \frac{n}{2 n + 1}$ , $n \in Z^{+}$ .

b) Hence, show that the sum of the first $(n + 1)$ terms of the series $\frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + \dots$ is $\frac{( n + 1 )}{( 2 n + 3 )}$ .

View solution

a) Let $P (n)$ : $r = 1 \sum n \frac{1}{( 2 r - 1 ) ( 2 r + 1 )} = \frac{n}{2 n + 1}$ .

Base case $n = 1$ : the left-hand side is $\frac{1}{1 \cdot 3} = \frac{1}{3}$ and the right-hand side is $\frac{1}{3}$ . So $P (1)$ holds.
Inductive hypothesis: suppose $P (k)$ is true for some $k \in Z^{+}$ , that is, $r = 1 \sum k \frac{1}{( 2 r - 1 ) ( 2 r + 1 )} = \frac{k}{2 k + 1}$ .
For $n = k + 1$ , we add the next term: $r = 1 \sum k + 1 \frac{1}{( 2 r - 1 ) ( 2 r + 1 )} = \frac{k}{2 k + 1} + \frac{1}{( 2 k + 1 ) ( 2 k + 3 )} = \frac{k ( 2 k + 3 ) + 1}{( 2 k + 1 ) ( 2 k + 3 )} .$
Expanding the numerator: $2 k^{2} + 3 k + 1 = (2 k + 1) (k + 1)$ , so $\frac{( 2 k + 1 ) ( k + 1 )}{( 2 k + 1 ) ( 2 k + 3 )} = \frac{k + 1}{2 k + 3} = \frac{k + 1}{2 ( k + 1 ) + 1},$ which is $P (k + 1)$ .
As $P (1)$ is true and $P (k) \Rightarrow P (k + 1)$ , by the principle of mathematical induction $P (n)$ is true for all $n \in Z^{+}$ .

b) 1. The denominators $3 = 1 \cdot 3, 15 = 3 \cdot 5, 35 = 5 \cdot 7, 63 = 7 \cdot 9, \dots$ show that the $r$ th term is $\frac{1}{( 2 r - 1 ) ( 2 r + 1 )}$ . 2. The sum of the first $n + 1$ terms is that of part a) with $n + 1$ in place of $n$ : $\frac{n + 1}{2 ( n + 1 ) + 1} = \frac{n + 1}{2 n + 3} .$

Q.E.D.

Arithmetic sequences

Exercise 9 BasicIB · P1 May 2002

Consider the arithmetic series $2, 5, 8, \dots$

a) Find an expression for $S_{n}$ , the sum of the first $n$ terms.

b) Find the value of $n$ for which $S_{n} = 1365$ .

Show answer

Answer: a) $S_{n} = \frac{3 n ^{2} + n}{2}$ ; b) $n = 30$

Show working and answer

a) 1. It is arithmetic with $u_{1} = 2$ and $d = 3$ .

$S_{n} = \frac{n}{2} [2 u_{1} + (n - 1) d] = \frac{n}{2} [4 + 3 (n - 1)] = \frac{n}{2} (3 n + 1)$ .

b) 1. We equate: $\frac{n ( 3 n + 1 )}{2} = 1365$ , so $3 n^{2} + n - 2730 = 0$ .

Discriminant: $1 + 4 \cdot 3 \cdot 2730 = 32761 = 18 1^{2}$ .
$n = \frac{- 1 \pm 181}{6}$ , that is, $n = 30$ or $n = - \frac{91}{3}$ .
We reject the negative value (it is not a positive integer). Therefore $n = 30$ .

Answer: a) $S_{n} = \frac{3 n ^{2} + n}{2}$ ; b) $n = 30$

Exercise 10 BasicIB · P1 TZ0 May 2006

In an arithmetic sequence, the second term is $7$ and the sum of the first five terms is $50$ . Find the common difference of this arithmetic sequence.

Show answer

Answer: $d = 3$

Show working and answer

Let $u_{1}$ be the first term and $d$ the common difference. The second term satisfies $u_{1} + d = 7$ .
The sum of the first five terms is $S_{5} = \frac{5}{2} (2 u_{1} + 4 d) = 5 (u_{1} + 2 d) = 50$ , so $u_{1} + 2 d = 10$ .
Subtracting the first equation from the second: $(u_{1} + 2 d) - (u_{1} + d) = 10 - 7$ , that is, $d = 3$ .
Check: $u_{1} = 4$ , and the terms $4, 7, 10, 13, 16$ add up to $50$ .

Answer: $d = 3$

Exercise 11 BasicIB · P1 May 1999

The second term of an arithmetic sequence is $7$ . The sum of the first four terms of the arithmetic sequence is $12$ . Find the first term, $a$ , and the common difference, $d$ , of the sequence.

Show answer

Answer: $a = 15$ , $d = - 8$

Show working and answer

Second term: $a + d = 7$ .
Sum of the first four terms: $S_{4} = \frac{4}{2} (2 a + 3 d) = 12 \Rightarrow 2 a + 3 d = 6$ .
From the first equation, $a = 7 - d$ . Substituting: $14 - 2 d + 3 d = 6 \Rightarrow d = - 8$ .
Then $a = 7 - (- 8) = 15$ .
Check: the terms are $15, 7, - 1, - 9$ , and their sum is $12$ .

Answer: $a = 15$ , $d = - 8$

Exercise 12 BasicIB · P1 TZ1 May 2016

The fifth term of an arithmetic sequence is equal to $6$ and the sum of the first $12$ terms is $45$ . Find the first term and the common difference.

Show answer

Answer: $d = - \frac{3}{2}$ , $u_{1} = 12$

Show working and answer

Fifth term: $a + 4 d = 6$ .
Sum of the first 12 terms: $S_{12} = \frac{12}{2} (2 a + 11 d) = 45 \Rightarrow 2 a + 11 d = \frac{15}{2}$ .
From the first, $a = 6 - 4 d$ . Substituting: $12 - 8 d + 11 d = \frac{15}{2} \Rightarrow 3 d = - \frac{9}{2} \Rightarrow d = - \frac{3}{2}$ .
Then $a = 6 - 4 (- \frac{3}{2}) = 12$ .

Answer: $d = - \frac{3}{2}$ , $u_{1} = 12$

Exercise 13 BasicIB · P1 November 1999

An arithmetic sequence has $5$ and $13$ as its first two terms, respectively.

a) Write down, in terms of $n$ , an expression for the $n$ th term, $a_{n}$ .

b) Find the number of terms of the sequence that are less than $400$ .

Show answer

Answer: a) $a_{n} = 8 n - 3$ ; b) there are $50$ terms less than $400$

Show working and answer

First term $5$ and common difference $d = 13 - 5 = 8$ .
General term: $a_{n} = 5 + (n - 1) \cdot 8 = 8 n - 3$ .

We require $8 n - 3 < 400 \Rightarrow 8 n < 403 \Rightarrow n < 50.375$ .
Since $n$ is a positive integer, $n = 1, 2, \dots, 50$ .
Check: $a_{50} = 397 < 400$ and $a_{51} = 405 > 400$ .

Answer: a) $a_{n} = 8 n - 3$ ; b) there are $50$ terms less than $400$

Exercise 14 BasicIB · P1 TZ2 May 2022

The $n$ th term of an arithmetic sequence is given by $u_{n} = 15 - 3 n$ .

a) State the value of the first term $(u_{1})$ . [1 mark]

b) Given that the $n$ th term of this sequence is $- 33$ , find the value of $n$ . [2 marks]

c) Find the common difference $(d)$ . [2 marks]

Show answer

Answer: a) $u_{1} = 12$ ; b) $n = 16$ ; c) $d = - 3$

Show working and answer

a) 1. Substituting $n = 1$ : $u_{1} = 15 - 3 (1) = 12$ .

b) 1. We set up $15 - 3 n = - 33$ .

Then $- 3 n = - 48$ , and $n = 16$ .

c) 1. The common difference is $d = u_{n + 1} - u_{n} = [15 - 3 (n + 1)] - (15 - 3 n) = - 3$ .

Answer: a) $u_{1} = 12$ ; b) $n = 16$ ; c) $d = - 3$

Exercise 15 MediumIB · P2 TZ2 May 2011

In an arithmetic sequence whose $n$ th term is $u_{n}$ , it is known that $u_{4} = 7$ and $u_{9} = 22$ . Find the minimum value of $n$ for which $u_{1} + u_{2} + u_{3} + \dots + u_{n} > 10000$ .

Show answer

Answer: $n = 83$

Show working and answer

Arithmetic sequence: $u_{n} = u_{1} + (n - 1) d$ . With $u_{4} = 7$ and $u_{9} = 22$ : $u_{9} - u_{4} = 5 d = 15 \Rightarrow d = 3.$
$u_{1} = u_{4} - 3 d = 7 - 9 = - 2$ .
Sum of the first $n$ terms: $S_{n} = \frac{n}{2} [2 u_{1} + (n - 1) d] = \frac{n}{2} [- 4 + 3 (n - 1)] = \frac{n ( 3 n - 7 )}{2}$ .
We require $S_{n} > 10000$ : $3 n^{2} - 7 n > 20000 ⟺ 3 n^{2} - 7 n - 20000 > 0$ .
Roots of $3 n^{2} - 7 n - 20000 = 0$ : $n = \frac{7 \pm 49 + 240000}{6} = \frac{7 \pm 240049}{6}$ . The positive root is $n \approx 82.82$ . Since the parabola opens upwards and $n > 0$ , the inequality holds for $n > 82.82$ .
Check with integers: $S_{82} = \frac{82 \cdot 239}{2} = 9799 < 10000$ and $S_{83} = \frac{83 \cdot 242}{2} = 10043 > 10000$ .

Answer: $n = 83$

Exercise 16 MediumIB · P1 TZ0 November 2008

A rope of length 81 metres is cut into $n$ pieces of increasing length, these lengths forming an arithmetic sequence with common difference $d$ metres. Given that the lengths of the shortest and the longest pieces are 1.5 metres and 7.5 metres respectively, find the values of $n$ and $d$ .

Show answer

Answer: $n = 18, d = \frac{6}{17} \approx 0.353$ m

Show working and answer

The pieces form an arithmetic sequence with $u_{1} = 1.5$ , last term $u_{n} = 7.5$ and sum $S_{n} = 81$ .
We use $S_{n} = \frac{n}{2} (u_{1} + u_{n})$ : $81 = \frac{n}{2} (1.5 + 7.5) = \frac{n}{2} \cdot 9 \Rightarrow n = 18.$
We use $u_{n} = u_{1} + (n - 1) d$ : $7.5 = 1.5 + 17 d \Rightarrow 17 d = 6 \Rightarrow d = \frac{6}{17} .$

Answer: $n = 18, d = \frac{6}{17} \approx 0.353$ m

Exercise 17 MediumIB · P1 TZ0 November 2010

The mean of the first ten terms of an arithmetic sequence is equal to 6. The mean of the first twenty terms of the arithmetic sequence is equal to 16. Find the value of the 15th term of the sequence.

Show answer

Answer: $u_{15} = 25$

Show working and answer

The mean of the first $10$ terms is $\frac{S _{10}}{10} = 6$ , so $S_{10} = 60$ . The mean of the first $20$ terms is $16$ , so $S_{20} = 320$ .
We use $S_{n} = \frac{n}{2} [2 u_{1} + (n - 1) d]$ : $S_{10} = 5 (2 u_{1} + 9 d) = 60 \Rightarrow 2 u_{1} + 9 d = 12,$ $S_{20} = 10 (2 u_{1} + 19 d) = 320 \Rightarrow 2 u_{1} + 19 d = 32.$
Subtracting the two equations: $10 d = 20 \Rightarrow d = 2$ . Then $2 u_{1} = 12 - 18 = - 6 \Rightarrow u_{1} = - 3$ .
$u_{15} = u_{1} + 14 d = - 3 + 28 = 25$ .

Answer: $u_{15} = 25$

Exercise 18 MediumIB · P2 TZ2 May 2012

The sum of the first 16 terms of an arithmetic sequence is 212, and the 5th term is 8.

a) Find the first term and the common difference.

b) Find the smallest value of $n$ for which the sum of the first $n$ terms is greater than 600.

View solution

$S_{16} = \frac{16}{2} (2 u_{1} + 15 d) = 212 \Rightarrow 2 u_{1} + 15 d = 26.5$ .
$u_{5} = u_{1} + 4 d = 8 \Rightarrow 2 u_{1} + 8 d = 16$ .
Subtracting: $7 d = 10.5 \Rightarrow d = 1.5$ . Then $u_{1} = 8 - 4 \cdot 1.5 = 2$ .

Answer a): $u_{1} = 2, d = 1.5$

$S_{n} = \frac{n}{2} [4 + 1.5 (n - 1)] = \frac{n}{2} (1.5 n + 2.5) = \frac{n ( 3 n + 5 )}{4}$ .
We require $S_{n} > 600$ : $3 n^{2} + 5 n - 2400 > 0$ .
Positive root: $n = \frac{- 5 + 25 + 28800}{6} = \frac{- 5 + 28825}{6} \approx 27.46$ . The inequality holds for $n > 27.46$ .
Check: $S_{27} = \frac{27 \cdot 86}{4} = 580.5 < 600$ and $S_{28} = \frac{28 \cdot 89}{4} = 623 > 600$ .

Answer b): $n = 28$

Exercise 19 MediumIB · P2 TZ0 November 2013

The fourth term of an arithmetic sequence is 34 and the tenth term is 76.

a) Find the first term and the common difference.

b) The sum of the first $n$ terms is greater than 5000. Find the least possible value of $n$ .

Show answer

Answer: $u_{1} = 13$ , $d = 7$

Answer: $n = 37$

Show working and answer

$u_{4} = a + 3 d = 34$ and $u_{10} = a + 9 d = 76$ .
Subtracting: $6 d = 42$ , so $d = 7$ .
Then $a = 34 - 3 \cdot 7 = 13$ .

Answer: $u_{1} = 13$ , $d = 7$

$S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [26 + 7 (n - 1)] = \frac{n ( 7 n + 19 )}{2}$ .
We impose $S_{n} > 5000$ : $7 n^{2} + 19 n - 10000 > 0$ .
The positive root is $n = \frac{- 19 + 1 9 ^{2} + 4 \cdot 7 \cdot 10000}{14} = \frac{- 19 + 280361}{14} \approx 36.46$ .
As $n$ is an integer, $n \geq 37$ . Check: $S_{36} = 18 \cdot 271 = 4878 < 5000$ and $S_{37} = \frac{37 \cdot 278}{2} = 5143 > 5000$ .

Answer: $n = 37$

Exercise 20 MediumIB · P1 November 2000

Calculate the sum of the positive terms of the arithmetic sequence $85, 78, 71, \dots$

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Answer: $559$

Show working and answer

Arithmetic sequence with $u_{1} = 85$ and $d = - 7$ : $u_{n} = 85 - 7 (n - 1) = 92 - 7 n$ .
Positive terms: $92 - 7 n > 0 \Rightarrow n < \frac{92}{7} \approx 13.14$ . So there are 13 positive terms, and $u_{13} = 92 - 91 = 1$ .
Sum: $S_{13} = \frac{13}{2} (u_{1} + u_{13}) = \frac{13}{2} (85 + 1) = 13 \cdot 43 = 559$ .

Answer: $559$

Exercise 21 MediumIB · P2 TZ2 May 2014

a) (i) Find the sum of all the integers between 10 and 200 that are divisible by 7.

(ii) Express the above sum using sigma notation. An arithmetic sequence has first term 1000 and common difference $- 6$ . The sum of the first $n$ terms of this sequence is negative.

b) Find the least value of $n$ .

Show answer

Answer: a) (i) $2835$ ; (ii) $n = 1 \sum 27 (7 n + 7)$ ; b) $n = 335$

Show working and answer

a) (i) 1. The multiples of 7 between 10 and 200 are $14, 21, \dots, 196$ , since $14 = 7 \cdot 2$ is the first one greater than 10 and $196 = 7 \cdot 28$ is the last one less than 200 ( $7 \cdot 29 = 203 > 200$ ).

It is an arithmetic sequence with $u_{1} = 14$ , $d = 7$ . From $196 = 14 + (n - 1) 7$ we obtain $n - 1 = 26$ , that is, $n = 27$ terms.
$S_{27} = \frac{27}{2} (14 + 196) = \frac{27}{2} \cdot 210 = 2835$ .

(ii) Each term is $7 k$ with $k$ from 2 to 28, so the sum is written $k = 2 \sum 28 7 k$ (equivalently $n = 1 \sum 27 (7 n + 7)$ ).

b) 1. With $u_{1} = 1000$ and $d = - 6$ : $S_{n} = \frac{n}{2} [2 \cdot 1000 + (n - 1) (- 6)] = \frac{n}{2} (2006 - 6 n) = n (1003 - 3 n)$ .

We impose $S_{n} < 0$ with $n > 0$ : $1003 - 3 n < 0$ , that is, $n > \frac{1003}{3} = 334.33 \dots$
As $n$ is an integer, the least value is $n = 335$ .

Answer: a) (i) $2835$ ; (ii) $n = 1 \sum 27 (7 n + 7)$ ; b) $n = 335$

Exercise 22 MediumIB · P1 November 2003

The first four terms of an arithmetic sequence are $2$ , $a - b$ , $2 a + b + 7$ and $a - 3 b$ , where $a$ and $b$ are constants. Find $a$ and $b$ .

Show answer

Answer: $a = 2$ , $b = - 3$

Show working and answer

In an arithmetic sequence the difference between consecutive terms is constant.
We equate the first and second differences: $(a - b) - 2 = (2 a + b + 7) - (a - b)$ , that is, $a - b - 2 = a + 2 b + 7$ , so $- 3 b = 9$ and $b = - 3$ .
We equate the second and third differences: $(2 a + b + 7) - (a - b) = (a - 3 b) - (2 a + b + 7)$ , that is, $a + 2 b + 7 = - a - 4 b - 7$ .
With $b = - 3$ : $a - 6 + 7 = - a + 12 - 7$ , so $a + 1 = - a + 5$ , $2 a = 4$ and $a = 2$ .
Check: the terms are $2, 5, 8, 11$ , with $d = 3$ .

Answer: $a = 2$ , $b = - 3$

Exercise 23 MediumIB · P1 TZ2 May 2007

Consider the arithmetic sequence $- 6, 1, 8, 15, \dots$

Find the least number of terms needed for the sum of the series to be greater than 10,000.

Show answer

Answer: $n = 55$

Show working and answer

It is an arithmetic sequence with $u_{1} = - 6$ and $d = 7$ .
The sum of the first $n$ terms is $S_{n} = \frac{n}{2} (2 (- 6) + (n - 1) 7) = \frac{n}{2} (7 n - 19)$ .
We impose $S_{n} > 10000$ : $\frac{n ( 7 n - 19 )}{2} > 10000$ , that is, $7 n^{2} - 19 n - 20000 > 0$ .
The positive root of $7 n^{2} - 19 n - 20000 = 0$ is $n = \frac{19 + 361 + 560 000}{14} = \frac{19 + 560 361}{14} \approx 54.83$ .
As $n$ is a positive integer and the parabola is increasing for large $n$ , we need $n \geq 55$ .
Check: $S_{54} = 27 \cdot 359 = 9693 < 10000$ and $S_{55} = \frac{55}{2} \cdot 366 = 10065 > 10000$ .

Answer: $n = 55$

Exercise 24 MediumIB · P2 TZ1 May 2010

Calculate the sum of all the three-digit natural numbers that are not divisible by $3$ .

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Answer: $329400$

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The three-digit numbers are $100, 101, \dots, 999$ : an arithmetic sequence of $900$ terms. Its sum is $\frac{900}{2} (100 + 999) = 450 \cdot 1099 = 494550$ .
The three-digit multiples of $3$ are $102, 105, \dots, 999$ . From $999 = 102 + (n - 1) \cdot 3$ we get $n = 300$ terms. Their sum is $\frac{300}{2} (102 + 999) = 150 \cdot 1101 = 165150$ .
The sum of the numbers not divisible by $3$ is the difference: $494550 - 165150 = 329400$ .

Answer: $329400$

Exercise 25 MediumIB · P2 TZ0 November 2011

In an arithmetic sequence the first term is $8$ and the common difference is $\frac{1}{4}$ . Given that the sum of the first $2 n$ terms is equal to the sum of the following $n$ terms, find $n$ .

Show answer

Answer: $n = 63$

Show working and answer

For this sequence, $S_{m} = \frac{m}{2} (2 \cdot 8 + (m - 1) \frac{1}{4}) = \frac{m}{2} \cdot \frac{64 + m - 1}{4} = \frac{m ( m + 63 )}{8}$ .
The $n$ terms following the first $2 n$ terms add up to $S_{3 n} - S_{2 n}$ . The condition $S_{2 n} = S_{3 n} - S_{2 n}$ is equivalent to $S_{3 n} = 2 S_{2 n}$ .
Substituting: $\frac{3 n ( 3 n + 63 )}{8} = 2 \cdot \frac{2 n ( 2 n + 63 )}{8}$ .
As $n \neq = 0$ , we simplify: $3 (3 n + 63) = 4 (2 n + 63)$ , that is, $9 n + 189 = 8 n + 252$ .
Hence $n = 63$ .

Answer: $n = 63$

Exercise 26 MediumIB · P1 TZ1 May 2009

Consider the set of numbers $a, 2 a, 3 a, \dots, na$ , where $a$ and $n$ are positive integers.

(i) Verify that the expression for the mean of this set is $\frac{a ( n + 1 )}{2}$ .

(ii) Let $a = 4$ . Calculate the least value of $n$ for which the sum of these numbers exceeds the mean by 100.

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Answer: $n = 8$

Show working and answer

(i) The mean is the sum of the terms divided by the number of terms.

The numbers form an arithmetic sequence with first term $a$ and difference $a$ . Their sum is $S_{n} = \frac{n}{2} (a + na) = \frac{na ( n + 1 )}{2}$ .
The mean is $\frac{S _{n}}{n} = \frac{a ( n + 1 )}{2}$ . This is verified.

(ii) With $a = 4$ :

Sum: $S_{n} = \frac{4 n ( n + 1 )}{2} = 2 n (n + 1)$ . Mean: $\frac{4 ( n + 1 )}{2} = 2 (n + 1)$ .
Condition: $S_{n} - mean > 100$ , that is, $2 n (n + 1) - 2 (n + 1) > 100$ .
Factorising: $2 (n + 1) (n - 1) > 100 \Rightarrow n^{2} - 1 > 50 \Rightarrow n^{2} > 51$ .
As $7^{2} = 49 < 51$ and $8^{2} = 64 > 51$ , the least integer is $n = 8$ . (With $n = 7$ : $2 \cdot 8 \cdot 6 = 96 < 100$ ; with $n = 8$ : $2 \cdot 9 \cdot 7 = 126 > 100$ .)

Answer: $n = 8$

Exercise 27 MediumIB · P2 TZ0 November 2012

Find the sum of all the multiples of $3$ between $100$ and $500$ .

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Answer: $39900$

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The multiples of 3 form an arithmetic sequence with difference $3$ . The first one greater than 100 is $102$ ( $= 3 \cdot 34$ ) and the last one less than 500 is $498$ ( $= 3 \cdot 166$ ).
Number of terms: $498 = 102 + (n - 1) \cdot 3 \Rightarrow n - 1 = 132 \Rightarrow n = 133$ .
Sum: $S = \frac{n}{2} (u_{1} + u_{n}) = \frac{133}{2} (102 + 498) = \frac{133}{2} \cdot 600 = 133 \cdot 300 = 39900$ .

Answer: $39900$

Exercise 28 MediumIB · P1 November 1996

Calculate the sum of all the integers between $200$ and $500$ that are divisible by $7$ .

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Answer: $15050$

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The first multiple of $7$ greater than $200$ : as $200/7 \approx 28.6$ , it is $7 \cdot 29 = 203$ .
The last multiple of $7$ less than $500$ : as $500/7 \approx 71.4$ , it is $7 \cdot 71 = 497$ .
The multiples form an arithmetic sequence with first term $u_{1} = 203$ , difference $d = 7$ and last term $497$ . We find the number of terms from $u_{n} = u_{1} + (n - 1) d$ : $497 = 203 + 7 (n - 1)$ , so $n - 1 = 42$ and $n = 43$ .
Sum of an arithmetic sequence: $S_{n} = \frac{n}{2} (u_{1} + u_{n})$ , so $S_{43} = \frac{43}{2} (203 + 497) = \frac{43}{2} \cdot 700 = 43 \cdot 350 = 15050$ .

Answer: $15050$

Exercise 29 MediumIB · P1 TZ2 May 2019

In an arithmetic sequence the sum of the $3^{\circ}$ and $8^{\circ}$ terms is equal to $1$ . Given that the sum of the first seven terms is $35$ , determine the first term and the common difference.

Show answer

Answer: $u_{1} = 14$ and $d = - 3$

Show working and answer

Let $a$ be the first term and $d$ the common difference.

Condition on the $3^{\circ}$ and $8^{\circ}$ terms: $(a + 2 d) + (a + 7 d) = 1$ , that is, $2 a + 9 d = 1$ . (1)
Sum of the first seven terms, $S_{7} = \frac{7}{2} (2 a + 6 d) = 35$ . Dividing by $7$ and multiplying by $2$ : $2 a + 6 d = 10$ . (2)
Subtracting (2) from (1): $3 d = 1 - 10 = - 9$ , so $d = - 3$ .
Substituting in (2): $2 a - 18 = 10$ , so $a = 14$ .
Check in (1): $2 \cdot 14 + 9 \cdot (- 3) = 28 - 27 = 1$ . Correct.

Answer: $u_{1} = 14$ and $d = - 3$

Exercise 30 MediumIB · P2 TZ1 May 2018

The third term of an arithmetic sequence is $1407$ and the tenth term is $1183$ .

a) Find the first term and the common difference of the sequence.

b) Calculate the number of positive terms in the sequence.

Show answer

Answer: a) $u_{1} = 1471$ and $d = - 32$ ; b) $46$ positive terms

Show working and answer

a) 1. Using $u_{n} = a + (n - 1) d$ : $u_{3} = a + 2 d = 1407$ and $u_{10} = a + 9 d = 1183$ . 2. Subtracting: $7 d = 1183 - 1407 = - 224$ , so $d = - 32$ . 3. Substituting: $a = 1407 - 2 (- 32) = 1407 + 64 = 1471$ .

b) 1. We require $u_{n} > 0$ : $1471 - 32 (n - 1) > 0$ , so $n - 1 < \frac{1471}{32} = 45.97 \dots$ 2. Therefore $n < 46.97$ , so $n \leq 46$ . 3. Check: $u_{46} = 1471 - 32 \cdot 45 = 31 > 0$ and $u_{47} = 1471 - 32 \cdot 46 = - 1 < 0$ . 4. There are $46$ positive terms.

Answer: a) $u_{1} = 1471$ and $d = - 32$ ; b) $46$ positive terms

Exercise 31 MediumIB · P1 TZ2 May 2016

Consider the expansion of $(1 + x)^{n}$ in ascending powers of $x$ , where $n \geq 3$ .

a) Write down the first four terms of the expansion. The coefficients of the second, third and fourth terms of the expansion are consecutive terms of an arithmetic sequence.

b) (i) Show that $n^{3} - 9 n^{2} + 14 n = 0$ .

(ii) Hence, find the value of $n$ .

Show answer

Answer: a) $(0 n) + (1 n) x + (2 n) x^{2} + (3 n) x^{3}$ ; b) (ii) $n = 7$

Show working and answer

a) By the binomial theorem: $(1 + x)^{n} = (0 n) + (1 n) x + (2 n) x^{2} + (3 n) x^{3} + \dots$ with coefficients $1, n, \frac{n ( n - 1 )}{2}, \frac{n ( n - 1 ) ( n - 2 )}{6}$ .

b) (i)

The coefficients of the second, third and fourth terms are $(1 n), (2 n), (3 n)$ . If they are in arithmetic sequence, the middle one is the mean of the outer two: $2 (2 n) = (1 n) + (3 n)$ .
Substituting: $n (n - 1) = n + \frac{n ( n - 1 ) ( n - 2 )}{6}$ .
Multiplying by 6: $6 n^{2} - 6 n = 6 n + n (n^{2} - 3 n + 2) = 6 n + n^{3} - 3 n^{2} + 2 n$ .
Moving everything to one side: $0 = n^{3} - 9 n^{2} + 14 n$ .

Q.E.D.

(ii)

Factorising: $n (n^{2} - 9 n + 14) = n (n - 2) (n - 7) = 0$ .
Solutions: $n = 0, 2, 7$ . Since $n \geq 3$ , only $n = 7$ is valid.

Answer: a) $(0 n) + (1 n) x + (2 n) x^{2} + (3 n) x^{3}$ ; b) (ii) $n = 7$

Exercise 32 MediumIB · P2 May 1999

The ratio of the fifth term to the twelfth term of an arithmetic sequence is $\frac{6}{13}$ . If every term of this sequence is positive, and the product of the first term and the third term is $32$ , find the sum of the first $100$ terms of this sequence.

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Answer: $S_{100} = 10300$

Show working and answer

Ratio of terms: $\frac{a + 4 d}{a + 11 d} = \frac{6}{13} \Rightarrow 13 a + 52 d = 6 a + 66 d \Rightarrow 7 a = 14 d \Rightarrow a = 2 d$ .
Product of the first and third terms: $a (a + 2 d) = 32$ . Substituting $a = 2 d$ : $2 d \cdot 4 d = 8 d^{2} = 32 \Rightarrow d^{2} = 4 \Rightarrow d = \pm 2$ .
If $d = - 2$ , then $a = - 4$ and the terms would be negative, which is not valid (all the terms are positive). We discard this option.
With $d = 2$ : $a = 4$ . All the terms are positive.
Sum of the first 100 terms: $S_{100} = \frac{100}{2} (2 \cdot 4 + 99 \cdot 2) = 50 \cdot 206 = 10300$ .

Answer: $S_{100} = 10300$

Exercise 33 MediumIB · P1 TZ0 November 2016

The sum of the first $n$ terms of a sequence ${u_{n}}$ is given by $S_{n} = 3 n^{2} - 2 n$ , where $n \in Z^{+}$ .

a) Write down the value of $u_{1}$ .

b) Find the value of $u_{6}$ .

c) Show that ${u_{n}}$ is an arithmetic sequence, clearly stating the common difference of the sequence.

Show answer

Answer: a) $u_{1} = 1$ ; b) $u_{6} = 31$ ; c) $d = 6$

Show working and answer

a) $u_{1} = S_{1} = 3 \cdot 1^{2} - 2 \cdot 1 = 1$ .

$u_{6} = S_{6} - S_{5}$ .
$S_{6} = 3 \cdot 36 - 12 = 96$ and $S_{5} = 3 \cdot 25 - 10 = 65$ .
$u_{6} = 96 - 65 = 31$ .

For $n \geq 2$ : $u_{n} = S_{n} - S_{n - 1} = 3 n^{2} - 2 n - [3 (n - 1)^{2} - 2 (n - 1)]$ .
Expanding: $3 n^{2} - 2 n - 3 n^{2} + 6 n - 3 + 2 n - 2 = 6 n - 5$ .
For $n = 1$ , $6 \cdot 1 - 5 = 1 = u_{1}$ , so the formula $u_{n} = 6 n - 5$ holds for all $n \geq 1$ .
We compute $u_{n + 1} - u_{n} = 6 (n + 1) - 5 - (6 n - 5) = 6$ , which is constant and independent of $n$ .
A sequence with a constant difference between consecutive terms is an arithmetic sequence, with common difference $d = 6$ .

Q.E.D.

Answer: a) $u_{1} = 1$ ; b) $u_{6} = 31$ ; c) $d = 6$

Exercise 34 MediumIB · P2 TZ1 May 2004

Find an expression for the sum of the first $35$ terms of the series $ln x^{2} + ln \frac{x ^{2}}{y} + ln \frac{x ^{2}}{y ^{2}} + ln \frac{x ^{2}}{y ^{3}} + \dots$ giving your answer in the form $ln \frac{x ^{m}}{y ^{n}}$ , where $m, n \in N$ .

Show answer

Answer: $S_{35} = ln (\frac{x ^{70}}{y ^{595}})$

Show working and answer

We use $ln \frac{x ^{2}}{y ^{k}} = 2 ln x - k ln y$ . The terms are $2 ln x, 2 ln x - ln y, 2 ln x - 2 ln y, \dots$
It is an arithmetic sequence with $u_{1} = 2 ln x$ and difference $d = - ln y$ .
Sum of the first 35 terms: $S_{35} = \frac{35}{2} [2 \cdot 2 ln x + 34 (- ln y)] = \frac{35}{2} (4 ln x - 34 ln y)$ .
Expanding: $S_{35} = 70 ln x - 595 ln y$ .
By the properties of logarithms ( $a ln b = ln b^{a}$ and $ln A - ln B = ln \frac{A}{B}$ ): $S_{35} = ln x^{70} - ln y^{595} = ln \frac{x ^{70}}{y ^{595}}$ .

Answer: $S_{35} = ln (\frac{x ^{70}}{y ^{595}})$

Exercise 35 MediumIB · P2 TZ2 May 2021

In an arithmetic sequence, the first term is $60$ and the common difference is $- 2.5$ .

a) Given that the $k$ th term of the sequence is zero, find the value of $k$ . Let $S_{n}$ be the sum of the first $n$ terms of the sequence.

b) Find the maximum value of $S_{n}$ .

Show answer

Answer: a) $k = 25$ ; b) $S_{m a x} = 750$

Show working and answer

a) The general term is $u_{n} = 60 + (n - 1) (- 2.5)$ .

$u_{k} = 0 \Rightarrow 60 - 2.5 (k - 1) = 0 \Rightarrow k - 1 = 24$ .
$k = 25$ .

b) $S_{n} = \frac{n}{2} [2 \cdot 60 + (n - 1) (- 2.5)]$ . The terms are positive up to $u_{24} = 2.5$ , $u_{25} = 0$ , and negative after that, so the sum stops growing at $n = 25$ (with $n = 24$ it is the same value).

$S_{25} = \frac{25}{2} (120 - 60) = 750$ .
Check: $S_{24} = 12 (120 - 57.5) = 750$ .

Answer: a) $k = 25$ ; b) $S_{m a x} = 750$

Exercise 36 MediumIB · P1 TZ1 May 2021

Consider the arithmetic sequence for which $u_{8} = S_{8} = 8$ . Calculate the value of the first term, $u_{1}$ , and the value of the common difference, $d$ .

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Answer: $u_{1} = - 6$ , $d = 2$

Show working and answer

$u_{8} = u_{1} + 7 d = 8$ . (1)
$S_{8} = \frac{8}{2} (2 u_{1} + 7 d) = 8 \Rightarrow 2 u_{1} + 7 d = 2$ . (2)
Subtracting (1) from (2): $u_{1} = 2 - 8 = - 6$ .
Substituting into (1): $- 6 + 7 d = 8 \Rightarrow d = 2$ .

Answer: $u_{1} = - 6$ , $d = 2$

Exercise 37 HardIB · P1 TZ1 May 2013

The first terms of an arithmetic sequence are $\frac{1}{lo g _{2} x}, \frac{1}{lo g _{8} x}, \frac{1}{lo g _{32} x}, \frac{1}{lo g _{128} x}, \dots$

Calculate $x$ if the sum of the first $20$ terms of the sequence is equal to $100$ .

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Answer: $x = 16$

Show working and answer

Using the change of base $\frac{1}{lo g _{b} x} = lo g_{x} b$ . The terms are $lo g_{x} 2, lo g_{x} 8, lo g_{x} 32, lo g_{x} 128, \dots$
As $8 = 2^{3}$ , $32 = 2^{5}$ , $128 = 2^{7}$ , and $lo g_{x} 2^{k} = k lo g_{x} 2$ , the terms are $L, 3 L, 5 L, 7 L, \dots$ with $L = lo g_{x} 2$ .
It is an arithmetic sequence with $u_{1} = L$ and $d = 2 L$ .
Sum of the first $20$ : $S_{20} = \frac{20}{2} (2 L + 19 \cdot 2 L) = 10 \cdot 40 L = 400 L$ .
Equating: $400 L = 100$ , so $L = \frac{1}{4}$ , that is, $lo g_{x} 2 = \frac{1}{4}$ .
Writing in exponential form, $x^{1/4} = 2$ , so $x = 2^{4} = 16$ .
Check: $lo g_{2} 16 = 4$ , so the first term is $\frac{1}{4} = L$ . Correct.

Answer: $x = 16$

Geometric sequences

Exercise 38 BasicIB · P1 May 2001

The $n$ th term, $u_{n}$ , of a geometric sequence is given by $u_{n} = 3 (4)^{n + 1}$ , $n \in Z^{+}$ .

a) Calculate the common ratio $r$ .

b) Calculate $S_{n}$ , the sum of the first $n$ terms of the sequence.

Show answer

Answer: a) $r = 4$ ; b) $S_{n} = 16 (4^{n} - 1)$

Show working and answer

a) 1. The ratio is the quotient of consecutive terms: $r = \frac{u _{n + 1}}{u _{n}} = \frac{3 ( 4 ) ^{n + 2}}{3 ( 4 ) ^{n + 1}} = 4$ .

b) 1. First term: $u_{1} = 3 (4)^{2} = 48$ .

$S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1} = \frac{48 ( 4 ^{n} - 1 )}{3} = 16 (4^{n} - 1)$ .

Answer: a) $r = 4$ ; b) $S_{n} = 16 (4^{n} - 1)$

Exercise 39 MediumIB · P2 TZ0 November 2008

A geometric sequence has first term 2 and common ratio 1.05. Find the value of the smallest term greater than 500.

Show answer

Answer: $u_{115} = 520.7$

Show working and answer

The $n$ th term: $u_{n} = 2 \cdot 1.0 5^{n - 1}$ .
We require $u_{n} > 500$ : $1.0 5^{n - 1} > 250$ .
Taking logarithms (the base $1.05 > 1$ , so the inequality sign is preserved): $(n - 1) ln 1.05 > ln 250 \Rightarrow n - 1 > \frac{ln 250}{ln 1.05} \approx \frac{5.5215}{0.048790} \approx 113.17.$
So $n - 1 \geq 114$ , i.e. $n \geq 115$ . The smallest term greater than $500$ is $u_{115}$ .
$u_{115} = 2 \cdot 1.0 5^{114} \approx 520.7$ . Check: $u_{114} = 2 \cdot 1.0 5^{113} \approx 495.9 < 500$ .

Answer: $u_{115} = 520.7$

Exercise 40 MediumIB · P1 TZ0 November 2013

The sum of the first two terms of a geometric series is equal to 10, while the sum of the first four terms is equal to 30.

a) Verify that the common ratio $r$ satisfies $r^{2} = 2$ .

b) Given that $r = 2$ ,

(i) Find the first term.

(ii) Find the sum of the first ten terms.

Show answer

Answer: a) $r^{2} = 2$ ; b) (i) $a = \frac{10}{1 + 2} = 10 (2 - 1)$ ; (ii) $S_{10} = 310$

Show working and answer

a) 1. Let $a$ be the first term. The sum of the first two terms is $S_{2} = a + a r = 10$ .

The sum of the first four terms is $S_{4} = a + a r + a r^{2} + a r^{3} = (a + a r) + r^{2} (a + a r)$ , that is, $S_{4} = S_{2} (1 + r^{2})$ .
Substituting: $30 = 10 (1 + r^{2})$ , so $1 + r^{2} = 3$ and hence $r^{2} = 2$ .

b) (i) 1. With $r = 2$ , from $a (1 + r) = 10$ we obtain $a = \frac{10}{1 + 2}$ .

Rationalising by multiplying by the conjugate: $a = \frac{10 ( 2 - 1 )}{( 2 + 1 ) ( 2 - 1 )} = \frac{10 ( 2 - 1 )}{2 - 1} = 10 (2 - 1)$ .

(ii) 1. We use $S_{10} = \frac{a ( r ^{10} - 1 )}{r - 1}$ .

As $r^{2} = 2$ , $r^{10} = (r^{2})^{5} = 2^{5} = 32$ .
$S_{10} = \frac{10 ( 2 - 1 ) \cdot 31}{2 - 1} = 310$ .

Answer: a) $r^{2} = 2$ ; b) (i) $a = \frac{10}{1 + 2} = 10 (2 - 1)$ ; (ii) $S_{10} = 310$

Exercise 41 MediumIB · P1 TZ0 November 2007

The first and fourth terms of a geometric series are $18$ and $- \frac{2}{3}$ , respectively. Find:

a) the sum of the first $n$ terms of the series.

b) the sum of the infinitely many terms of the series.

Show answer

Answer: a) $S_{n} = \frac{27}{2} (1 - (- \frac{1}{3})^{n})$ ; b) $S_{\infty} = \frac{27}{2}$

Show working and answer

Let $u_{1} = 18$ and $u_{4} = - \frac{2}{3}$ .

In a geometric sequence $u_{4} = u_{1} r^{3}$ , so $r^{3} = \frac{- 2/3}{18} = - \frac{1}{27}$ and $r = - \frac{1}{3}$ .

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

$S_{n} = \frac{18 ( 1 - ( - \frac{1}{3} ) ^{n} )}{1 + \frac{1}{3}} = \frac{18 ( 1 - ( - \frac{1}{3} ) ^{n} )}{\frac{4}{3}} = \frac{27}{2} (1 - (- \frac{1}{3})^{n})$ .

b) As $∣ r ∣ = \frac{1}{3} < 1$ , the series converges and $S_{\infty} = \frac{u _{1}}{1 - r}$ .

$S_{\infty} = \frac{18}{1 + \frac{1}{3}} = \frac{18}{\frac{4}{3}} = \frac{27}{2}$ .

Answer: a) $S_{n} = \frac{27}{2} (1 - (- \frac{1}{3})^{n})$ ; b) $S_{\infty} = \frac{27}{2}$

Exercise 42 MediumIB · P2 TZ2 May 2019Proof

Suppose $u_{1}$ is the first term of a geometric series with common ratio $r$ . Prove by mathematical induction that the sum of the first $n$ terms, $S_{n}$ , is given by $S_{n} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ , where $n \in Z^{+}$ .

View solution

We want to prove $P (n)$ : $S_{n} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ for all $n \in Z^{+}$ (with $r \neq = 1$ ).

Base case $n = 1$ . $S_{1} = u_{1}$ and $\frac{u _{1} ( 1 - r )}{1 - r} = u_{1}$ . So $P (1)$ holds.
Induction hypothesis. Suppose $P (k)$ is true for some $k \in Z^{+}$ : $S_{k} = \frac{u _{1} ( 1 - r ^{k} )}{1 - r}$ .
Inductive step. As $S_{k + 1} = S_{k} + u_{k + 1}$ and $u_{k + 1} = u_{1} r^{k}$ : $S_{k + 1} = \frac{u _{1} ( 1 - r ^{k} )}{1 - r} + u_{1} r^{k} = \frac{u _{1} ( 1 - r ^{k} ) + u _{1} r ^{k} ( 1 - r )}{1 - r} .$
Expanding the numerator: $u_{1} - u_{1} r^{k} + u_{1} r^{k} - u_{1} r^{k + 1} = u_{1} (1 - r^{k + 1})$ . Therefore $S_{k + 1} = \frac{u _{1} ( 1 - r ^{k + 1} )}{1 - r},$ which is $P (k + 1)$ .
As $P (1)$ is true and $P (k) \Rightarrow P (k + 1)$ , by induction $P (n)$ is true for all $n \in Z^{+}$ .

Hence proved.

Exercise 43 MediumIB · P2 TZ0 November 2012

A 1-metre metal bar is cut into 10 pieces, whose lengths form a geometric sequence. The longest piece has a length equal to 8 times the length of the shortest piece. Find the length of the shortest piece, rounded to the nearest millimetre.

Show answer

Answer: $29 mm$

Show working and answer

The lengths form a geometric sequence of 10 terms. Let $u_{1}$ be the shortest piece and $r$ the common ratio ( $r > 1$ ). The longest is $u_{10} = u_{1} r^{9} = 8 u_{1}$ , so $r^{9} = 8$ and $r = 8^{1/9} = 2^{1/3} \approx 1.2599$ .
The sum of the pieces is 1 m: $u_{1} \frac{r ^{10} - 1}{r - 1} = 1$ .
$r^{10} = 2^{10/3} \approx 10.0794$ , so $\frac{r ^{10} - 1}{r - 1} \approx \frac{9.0794}{0.2599} \approx 34.93$ .
$u_{1} \approx \frac{1}{34.93} \approx 0.02863 m \approx 28.6 mm$ .
Rounding to the nearest millimetre: $29$ mm.

Answer: $29 mm$

Exercise 44 MediumIB · P2 TZ0 November 2018

Consider a geometric sequence where the first term is $4$ and the fourth term is $- 2.916$ .

a) Find the common ratio of this sequence.

b) Find the sum to infinity of this sequence.

Show answer

Answer: a) $r = - 0.9$ ; b) $S_{\infty} = \frac{40}{19} \approx 2.11$

Show working and answer

a) 1. In a geometric sequence $u_{4} = u_{1} r^{3}$ , so $4 r^{3} = - 2.916$ . 2. $r^{3} = - 0.729$ , so $r = 3 - 0.729 = - 0.9$ .

b) 1. As $∣ r ∣ = 0.9 < 1$ , the sum converges: $S_{\infty} = \frac{u _{1}}{1 - r}$ . 2. $S_{\infty} = \frac{4}{1 - ( - 0.9 )} = \frac{4}{1.9} = \frac{40}{19} \approx 2.11$ .

Answer: a) $r = - 0.9$ ; b) $S_{\infty} = \frac{40}{19} \approx 2.11$

Exercise 45 MediumIB · P2 TZ0 November 2022

A geometric sequence has first term $50$ and fourth term $86.4$ . Let $S_{n}$ be the sum of the first $n$ terms of the sequence. Find the smallest value of $n$ for which $S_{n} > 33500$ .

Show answer

Answer: $n = 27$

Show working and answer

Geometric sequence: $u_{4} = u_{1} r^{3}$ , so $86.4 = 50 r^{3}$ and $r^{3} = 1.728$ , hence $r = 1.2$ .
Sum of the first $n$ terms: $S_{n} = \frac{50 ( 1. 2 ^{n} - 1 )}{1.2 - 1} = 250 (1. 2^{n} - 1)$ .
We require $250 (1. 2^{n} - 1) > 33500$ , that is, $1. 2^{n} > 135$ .
Taking logarithms (as $ln 1.2 > 0$ , the direction of the inequality is kept): $n > \frac{ln 135}{ln 1.2} \approx 26.90$ .
The smallest integer is $n = 27$ .

Answer: $n = 27$

Exercise 46 HardIB · P2 TZ0 November 2011Proof

If $a_{1}, a_{2}, a_{3}, \dots$ are terms of a geometric sequence with common ratio $r \neq = 1$ , use mathematical induction to verify that $(a_{1} - a_{2})^{2} + (a_{2} - a_{3})^{2} + (a_{3} - a_{4})^{2} + \dots + (a_{n} - a_{n + 1})^{2} = \frac{a _{1}^{2} ( 1 - r ) ( 1 - r ^{2 n} )}{1 + r}$

View solution

As the sequence is geometric, $a_{k} = a_{1} r^{k - 1}$ , so $a_{k} - a_{k + 1} = a_{1} r^{k - 1} (1 - r)$ and $(a_{k} - a_{k + 1})^{2} = a_{1}^{2} (1 - r)^{2} r^{2 k - 2}$ .

Let $P (n)$ : $k = 1 \sum n (a_{k} - a_{k + 1})^{2} = \frac{a _{1}^{2} ( 1 - r ) ( 1 - r ^{2 n} )}{1 + r}$ .

Base case $n = 1$ . Left-hand side: $(a_{1} - a_{2})^{2} = a_{1}^{2} (1 - r)^{2}$ . Right-hand side: $\frac{a _{1}^{2} ( 1 - r ) ( 1 - r ^{2} )}{1 + r} = \frac{a _{1}^{2} ( 1 - r ) ( 1 - r ) ( 1 + r )}{1 + r} = a_{1}^{2} (1 - r)^{2}$ . They coincide.
Hypothesis. Suppose $P (m)$ is true: $k = 1 \sum m (a_{k} - a_{k + 1})^{2} = \frac{a _{1}^{2} ( 1 - r ) ( 1 - r ^{2 m} )}{1 + r}$ .
Inductive step. The next term is $(a_{m + 1} - a_{m + 2})^{2} = a_{1}^{2} (1 - r)^{2} r^{2 m}$ . Then $k = 1 \sum m + 1 = \frac{a _{1}^{2} ( 1 - r ) ( 1 - r ^{2 m} )}{1 + r} + a_{1}^{2} (1 - r)^{2} r^{2 m} = \frac{a _{1}^{2} ( 1 - r )}{1 + r} [(1 - r^{2 m}) + (1 - r) (1 + r) r^{2 m}] .$
Inside the bracket: $1 - r^{2 m} + (1 - r^{2}) r^{2 m} = 1 - r^{2 m} + r^{2 m} - r^{2 m + 2} = 1 - r^{2 (m + 1)}$ .
Hence the sum is $\frac{a _{1}^{2} ( 1 - r ) ( 1 - r ^{2 (m + 1)} )}{1 + r}$ , which is $P (m + 1)$ .
As $P (1)$ is true and $P (m) \Rightarrow P (m + 1)$ , by induction the formula holds for all $n \in Z^{+}$ .

Hence proved.

Infinite sum of a geometric sequence

Exercise 47 BasicIB · P1 TZ1 May 2012

Calculate the value of $k$ if $r = 1 \sum \infty k (\frac{1}{3})^{r} = 7$

Show answer

Answer: $k = 14$

Show working and answer

The series is geometric with first term $k \cdot \frac{1}{3}$ and ratio $\frac{1}{3}$ ; as $∣ r ∣ < 1$ it converges.
$S_{\infty} = \frac{k /3}{1 - \frac{1}{3}} = \frac{k /3}{2/3} = \frac{k}{2}$ .
$\frac{k}{2} = 7 \Rightarrow k = 14$ .

Answer: $k = 14$

Exercise 48 MediumIB · P1 TZ2 May 2008

a) Find the sum to infinity of the geometric sequence: $27; - 9; 3; - 1; \dots$

b) Use mathematical induction to prove that for all $n \in Z^{+}$ the following holds $a + a r + a r^{2} + \dots + a r^{n - 1} = \frac{a ( 1 - r ^{n} )}{1 - r}$

View solution

First term $a = 27$ and common ratio $r = \frac{- 9}{27} = - \frac{1}{3}$ . Check: $- 9 \cdot (- \frac{1}{3}) = 3$ and $3 \cdot (- \frac{1}{3}) = - 1$ .
Since $∣ r ∣ = \frac{1}{3} < 1$ , the sum to infinity exists: $S_{\infty} = \frac{a}{1 - r} = \frac{27}{1 + \frac{1}{3}} = \frac{27}{\frac{4}{3}} = \frac{81}{4}$ .

Answer a): $S_{\infty} = \frac{81}{4} = 20.25$

b) We prove by induction that $P (n) : a + a r + \dots + a r^{n - 1} = \frac{a ( 1 - r ^{n} )}{1 - r}$ (with $r \neq = 1$ ).

Base case $n = 1$ : the left-hand side is $a$ ; the right-hand side is $\frac{a ( 1 - r )}{1 - r} = a$ . So $P (1)$ holds.
Inductive hypothesis: suppose $P (k)$ is true for some $k \in Z^{+}$ : $a + a r + \dots + a r^{k - 1} = \frac{a ( 1 - r ^{k} )}{1 - r} .$
Inductive step: for $n = k + 1$ , add the term $a r^{k}$ to both sides: $a + a r + \dots + a r^{k - 1} + a r^{k} = \frac{a ( 1 - r ^{k} )}{1 - r} + a r^{k} = \frac{a ( 1 - r ^{k} ) + a r ^{k} ( 1 - r )}{1 - r} .$
Expand the numerator: $a - a r^{k} + a r^{k} - a r^{k + 1} = a (1 - r^{k + 1})$ . Therefore $a + a r + \dots + a r^{k} = \frac{a ( 1 - r ^{k + 1} )}{1 - r},$ which is $P (k + 1)$ .
Since $P (1)$ is true and $P (k) \Rightarrow P (k + 1)$ , by the principle of mathematical induction $P (n)$ is true for all $n \in Z^{+}$ .

As required.

Exercise 49 MediumIB · P2 TZ2 May 2012

Each time a ball bounces, it reaches 95% of the height reached on the previous bounce. Initially the ball is dropped from a height of 4 metres.

a) What height does the ball reach after the fourth bounce?

b) How many times does the ball bounce before it no longer reaches a height of 1 metre?

c) What is the total distance travelled by the ball?

Show answer

Answer: $n = 28$

Answer: $156$ m

Show working and answer

The height after each bounce is multiplied by $0.95$ : this is a geometric sequence with first term $4$ (initial height) and common ratio $0.95$ .
After the fourth bounce: $4 \cdot 0.9 5^{4} = 3.258025 \dots \approx 3.26$ m.

After $n$ bounces the height is $4 \cdot 0.9 5^{n}$ . We require it to be less than 1: $4 \cdot 0.9 5^{n} < 1 \Rightarrow 0.9 5^{n} < 0.25 \Rightarrow n > \frac{ln 0.25}{ln 0.95} \approx 27.03.$
Check: after 27 bounces the height is $4 \cdot 0.9 5^{27} \approx 1.0015 > 1$ m, and after 28 bounces it is $\approx 0.951 < 1$ m.
Therefore, the smallest integer $n$ satisfying the condition is $28$ : at the 28th bounce the ball no longer reaches 1 m.

Answer: $n = 28$

Initial fall: $4$ m.
After that, at each bounce the ball goes up and comes down the same height. The rising heights are $3.8; 3.8 \cdot 0.95; \dots$ , an infinite geometric sequence with first term $3.8$ and common ratio $0.95$ , with sum $\frac{3.8}{1 - 0.95} = 76.$
Total distance: $4 + 2 \cdot 76 = 156$ m.

Answer: $156$ m

Exercise 50 MediumIB · P1 TZ2 May 2013

In a geometric sequence, the first term is $a$ , the common ratio is $r$ and the sum to infinity is 76. In a second geometric sequence, the first term is $a$ , the common ratio is $r^{3}$ and the sum to infinity is 36. Find $r$ .

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

Show working and answer

First sequence: $\frac{a}{1 - r} = 76$ . Second: $\frac{a}{1 - r ^{3}} = 36$ (with $∣ r ∣ < 1$ so that the sums exist).
Divide the first equation by the second to eliminate $a$ : $\frac{1 - r ^{3}}{1 - r} = \frac{76}{36} = \frac{19}{9} .$
Since $1 - r^{3} = (1 - r) (1 + r + r^{2})$ , this gives $1 + r + r^{2} = \frac{19}{9}$ , i.e. $9 r^{2} + 9 r - 10 = 0$ .
Factorise: $(3 r - 2) (3 r + 5) = 0$ , so $r = \frac{2}{3}$ or $r = - \frac{5}{3}$ .
The solution $r = - \frac{5}{3}$ is rejected because $∣ r ∣ > 1$ (the sum to infinity would not exist). For $r = \frac{2}{3}$ : $a = 76 \cdot \frac{1}{3} = \frac{76}{3}$ , and $\frac{a}{1 - r ^{3}} = \frac{76/3}{19/27} = 36$ . Correct.

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

Exercise 51 MediumIB · P1 November 2000

The sum of an infinite geometric series is $\frac{27}{2}$ , and the sum of its first three terms is 13. Calculate the first term.

Show answer

Answer: $u_{1} = 9$

Show working and answer

Infinite sum: $\frac{a}{1 - r} = \frac{27}{2}$ with $∣ r ∣ < 1$ .
Sum of three terms: $a \frac{1 - r ^{3}}{1 - r} = 13$ .
Dividing the second equation by the first: $1 - r^{3} = \frac{13}{27/2} = \frac{26}{27}$ , so $r^{3} = \frac{1}{27}$ and $r = \frac{1}{3}$ (which satisfies $∣ r ∣ < 1$ ).
From the first equation: $a = \frac{27}{2} (1 - \frac{1}{3}) = \frac{27}{2} \cdot \frac{2}{3} = 9$ .
Check: $9 + 3 + 1 = 13$ and $\frac{9}{1 - \frac{1}{3}} = \frac{27}{2}$ .

Answer: $u_{1} = 9$

Exercise 52 MediumIB · P1 November 2001

Consider the infinite geometric series $1 + (\frac{2 x}{3}) + (\frac{2 x}{3})^{2} + (\frac{2 x}{3})^{3} + \dots$

a) For which values of $x$ is the series convergent?

b) Calculate the sum of the series if $x = 1.2$ .

Show answer

Answer: a) $- \frac{3}{2} < x < \frac{3}{2}$ ; b) $S_{\infty} = 5$

Show working and answer

a) 1. It is a geometric series with first term $1$ and ratio $r = \frac{2 x}{3}$ .

It converges if and only if $∣ r ∣ < 1$ : $\frac{2 x}{3} < 1$ , that is, $∣ x ∣ < \frac{3}{2}$ .

b) 1. If $x = 1.2$ , $r = \frac{2 ( 1.2 )}{3} = 0.8$ , which satisfies $∣ r ∣ < 1$ .

$S_{\infty} = \frac{1}{1 - r} = \frac{1}{0.2} = 5$ .

Answer: a) $- \frac{3}{2} < x < \frac{3}{2}$ ; b) $S_{\infty} = 5$

Exercise 53 MediumIB · P1 May 2003

A geometric sequence has all its terms positive. The sum of the first two terms is 15 and the sum of the infinitely many terms of the sequence is 27. Find the value of:

a) the common ratio.

b) the first term

Show answer

Answer: a) $r = \frac{2}{3}$ ; b) $a = 9$

Show working and answer

Let $a$ be the first term and $r$ the common ratio. Data: $a + a r = 15$ and $\frac{a}{1 - r} = 27$ with $∣ r ∣ < 1$ .
From the second: $a = 27 (1 - r)$ . Substituting into the first: $27 (1 - r) (1 + r) = 15$ , so $1 - r^{2} = \frac{15}{27} = \frac{5}{9}$ and $r^{2} = \frac{4}{9}$ .
Thus $r = \pm \frac{2}{3}$ . As all the terms are positive and $a > 0$ , the ratio must be positive: $r = \frac{2}{3}$ (rejecting $- \frac{2}{3}$ ).
$a = 27 (1 - \frac{2}{3}) = 9$ .
Check: $9 + 6 = 15$ .

Answer: a) $r = \frac{2}{3}$ ; b) $a = 9$

Exercise 54 MediumIB · P1 TZ1 May 2004

A geometric sequence has a negative common ratio. The sum of the first two terms is $6$ . The infinite sum is $8$ . Calculate the common ratio and the first term.

Show answer

Answer: $r = - \frac{1}{2}, u_{1} = 12$

Show working and answer

Let $u_{1}$ be the first term and $r$ the common ratio. The sum of the first two terms is $u_{1} + u_{1} r = 6$ , that is, $u_{1} (1 + r) = 6$ .
The infinite sum is $\frac{u _{1}}{1 - r} = 8$ , so $u_{1} = 8 (1 - r)$ .
Substituting: $8 (1 - r) (1 + r) = 6$ , that is, $8 (1 - r^{2}) = 6$ , from which $1 - r^{2} = \frac{3}{4}$ and $r^{2} = \frac{1}{4}$ .
Thus $r = \pm \frac{1}{2}$ . As the ratio is negative, $r = - \frac{1}{2}$ (which satisfies $∣ r ∣ < 1$ , so the infinite sum exists).
Then $u_{1} = 8 (1 + \frac{1}{2}) = 12$ . Check: $12 + 12 (- \frac{1}{2}) = 6$ and $\frac{12}{1 + \frac{1}{2}} = 8$ .

Answer: $r = - \frac{1}{2}, u_{1} = 12$

Exercise 55 MediumIB · P1 TZ0 November 2006

The infinite sum of a geometric series is $32$ . The sum of the first four terms is $30$ and all the terms are positive. Find the difference between the infinite sum and the sum of the first eight terms.

Show answer

Answer: $S_{\infty} - S_{8} = \frac{1}{8}$

Show working and answer

Let $a$ be the first term and $r$ the common ratio. Infinite sum: $\frac{a}{1 - r} = 32$ .
Sum of the first four terms: $\frac{a ( 1 - r ^{4} )}{1 - r} = 30$ .
Dividing the second equation by the first: $1 - r^{4} = \frac{30}{32} = \frac{15}{16}$ , so $r^{4} = \frac{1}{16}$ .
Thus $r = \pm \frac{1}{2}$ ; as all the terms are positive, $r = \frac{1}{2}$ (with $a > 0$ ). Then $a = 32 (1 - \frac{1}{2}) = 16$ .
The required difference is $S_{\infty} - S_{8} = \frac{a}{1 - r} - \frac{a ( 1 - r ^{8} )}{1 - r} = \frac{a r ^{8}}{1 - r} = 32 r^{8}$ .
Then $32 (\frac{1}{2})^{8} = \frac{32}{256} = \frac{1}{8}$ .

Answer: $S_{\infty} - S_{8} = \frac{1}{8}$

Exercise 56 MediumIB · P1 TZ1 May 2007

An infinite geometric series is given by $k = 1 \sum \infty 2 (4 - 3 x)^{k}$ .

a) Calculate the values of $x$ for which the series has a finite sum.

b) For $x = 1.2$ , calculate the minimum number of terms needed for the sum to be greater than 1.328.

Show answer

Answer: a) $1 < x < \frac{5}{3}$ ; b) $n = 7$

Show working and answer

a) 1. It is a geometric series with ratio $r = 4 - 3 x$ and first term $2 (4 - 3 x)$ . 2. It has a finite sum if $∣ r ∣ < 1$ : $- 1 < 4 - 3 x < 1$ . 3. Subtracting $4$ : $- 5 < - 3 x < - 3$ . Dividing by $- 3$ (the inequalities reverse): $1 < x < \frac{5}{3}$ .

b) 1. For $x = 1.2$ : $r = 4 - 3.6 = 0.4$ and the first term is $u_{1} = 2 (0.4) = 0.8$ . 2. $S_{n} = \frac{0.8 ( 1 - 0. 4 ^{n} )}{1 - 0.4} = \frac{4}{3} (1 - 0. 4^{n})$ . 3. We impose $\frac{4}{3} (1 - 0. 4^{n}) > 1.328$ : $1 - 0. 4^{n} > 0.996$ , so $0. 4^{n} < 0.004$ . 4. Taking logarithms (and reversing the inequality when dividing by $ln 0.4 < 0$ ): $n > \frac{ln 0.004}{ln 0.4} \approx 6.03$ . 5. Check: $0. 4^{6} = 0.004096 > 0.004$ (not enough) and $0. 4^{7} = 0.0016384 < 0.004$ (enough).

Answer: a) $1 < x < \frac{5}{3}$ ; b) $n = 7$

Exercise 57 MediumIB · P2 TZ0 November 2011

a) Find the set of values of $x$ for which the series $n = 1 \sum \infty (\frac{2 x}{x + 1})^{n}$ has an infinite sum.

b) Hence, find that sum, in terms of $x$ .

Show answer

Answer: a) $- \frac{1}{3} < x < 1$ ; b) $S_{\infty} = \frac{2 x}{1 - x}$

Show working and answer

It is a geometric series with first term and ratio $r = \frac{2 x}{x + 1}$ (with $x \neq = - 1$ ). It has an infinite sum if and only if $∣ r ∣ < 1$ .
$\frac{2 x}{x + 1} < 1 ⟺ 4 x^{2} < (x + 1)^{2}$ (squaring, since both sides are non-negative).
$4 x^{2} < x^{2} + 2 x + 1 \Rightarrow 3 x^{2} - 2 x - 1 < 0 \Rightarrow (3 x + 1) (x - 1) < 0$ .
The parabola is negative between its roots: $- \frac{1}{3} < x < 1$ . This interval does not contain $x = - 1$ , so nothing else needs to be excluded.

First term: $u_{1} = \frac{2 x}{x + 1}$ and ratio $r = \frac{2 x}{x + 1}$ . Then $S_{\infty} = \frac{r}{1 - r}$ .
$1 - r = \frac{x + 1 - 2 x}{x + 1} = \frac{1 - x}{x + 1}$ .
$S_{\infty} = \frac{\frac{2 x}{x + 1}}{\frac{1 - x}{x + 1}} = \frac{2 x}{1 - x}$ .

Answer: a) $- \frac{1}{3} < x < 1$ ; b) $S_{\infty} = \frac{2 x}{1 - x}$

Exercise 58 MediumIB · P2 TZ2 May 2016

The sum of the second and third terms of a geometric sequence is equal to $96$ . The sum to infinity of this sequence is equal to $500$ . Find the possible values of the common ratio, $r$ .

Show answer

Answer: $r = 0.885$ (exactly $\frac{- 1 + 97}{10}$ ) and $r = 0.2$

Show working and answer

Let $a$ be the first term. The second and third terms are $a r + a r^{2} = 96$ .
Sum to infinity (which requires $∣ r ∣ < 1$ ): $\frac{a}{1 - r} = 500 \Rightarrow a = 500 (1 - r)$ .
Substituting: $500 (1 - r) r (1 + r) = 96 \Rightarrow 500 r (1 - r^{2}) = 96 \Rightarrow 125 r^{3} - 125 r + 24 = 0$ .
We try $r = \frac{1}{5}$ : $\frac{125}{125} - 25 + 24 = 0$ . It holds, so $(5 r - 1)$ is a factor: $(5 r - 1) (25 r^{2} + 5 r - 24) = 0.$
From $5 r - 1 = 0$ : $r = \frac{1}{5}$ .
From $25 r^{2} + 5 r - 24 = 0$ : $r = \frac{- 5 \pm 25 + 2400}{50} = \frac{- 1 \pm 97}{10}$ , that is, $r \approx 0.885$ or $r \approx - 1.085$ .
We discard $r \approx - 1.085$ because $∣ r ∣ > 1$ and the infinite sum would not exist. We check that $0.2$ and $0.885$ satisfy $∣ r ∣ < 1$ .

Answer: $r = 0.885$ (exactly $\frac{- 1 + 97}{10}$ ) and $r = 0.2$

Exercise 59 MediumIB · P2 May 1997

A rubber ball is dropped from a height of $81$ metres. Each time it hits the ground, it rebounds to two thirds of the distance from which it fell.

a) Find the maximum height of the ball between the fifth and sixth bounces.

b) What is the total distance travelled by the ball from the moment it is dropped until the moment it hits the ground for the sixth time?

c) Suppose the ball continues to bounce indefinitely. What is the total distance travelled by the ball?

Show answer

Answer: a) $\frac{32}{3} \approx 10.7$ m; b) $\frac{1087}{3} \approx 362.3$ m; c) $405$ m.

Show working and answer

The heights of the bounces form a geometric sequence with first term $81 \cdot \frac{2}{3} = 54$ and ratio $r = \frac{2}{3}$ .

a) The maximum height between the 5th and 6th bounces is the one reached after the 5th bounce, that is, the 5th term of this sequence:

$h_{5} = 81 (\frac{2}{3})^{5} = 81 \cdot \frac{32}{243} = \frac{32}{3} \approx 10.7$ m.

b) The ball falls $81$ m, and then, up to the sixth impact, it goes up and down after each of the first 5 bounces.

Sum of the 5 bounce heights (sum of a geometric sequence, $S_{n} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ ): $S_{5} = \frac{54 ( 1 - ( \frac{2}{3} ) ^{5} )}{1 - \frac{2}{3}} = 162 \cdot \frac{211}{243} = \frac{422}{3}$ .
Each height is travelled twice (up and down): $D = 81 + 2 \cdot \frac{422}{3} = 81 + \frac{844}{3} = \frac{1087}{3} \approx 362.3$ m.

c) If it bounces indefinitely, the ratio satisfies $∣ r ∣ < 1$ and the sum to infinity exists: $S_{\infty} = \frac{u _{1}}{1 - r}$ .

$S_{\infty} = \frac{54}{1 - \frac{2}{3}} = 162$ .
Total distance: $D = 81 + 2 \cdot 162 = 405$ m.

Answer: a) $\frac{32}{3} \approx 10.7$ m; b) $\frac{1087}{3} \approx 362.3$ m; c) $405$ m.

Exercise 60 MediumIB · P1 TZ1 May 2008

The common ratio of the terms in a geometric series is $2^{x}$ .

a) State the set of values of $x$ for which the sum to infinity of the series exists.

b) If the first term of the series is $35$ , find the value of $x$ for which the sum to infinity is $40$ .

Show answer

Answer: a) $x < 0$ ; b) $x = - 3$ .

Show working and answer

a) The sum to infinity of a geometric series exists if $∣ r ∣ < 1$ . Here $r = 2^{x} > 0$ , so it is enough that $2^{x} < 1$ .

$2^{x} < 1 = 2^{0}$ and, since the exponential function of base 2 is increasing, $x < 0$ .

b) With $u_{1} = 35$ and $S_{\infty} = \frac{u _{1}}{1 - r} = 40$ :

$\frac{35}{1 - 2 ^{x}} = 40 \Rightarrow 1 - 2^{x} = \frac{35}{40} = \frac{7}{8}$ .
$2^{x} = \frac{1}{8} = 2^{- 3} \Rightarrow x = - 3$ , which satisfies $x < 0$ .

Answer: a) $x < 0$ ; b) $x = - 3$ .

Exercise 61 MediumIB · P2 TZ0 November 2021

The sum of the first $n$ terms of a geometric sequence is given by $S_{n} = r = 1 \sum n \frac{2}{3} \cdot (\frac{7}{8})^{r}$

a) Find the first term of the sequence, $u_{1}$ .

b) Calculate $S_{\infty}$ .

c) Calculate the least value of $n$ for which $S_{\infty} - S_{n} < 0.001$ .

Show answer

Answer: a) $u_{1} = \frac{7}{12}$ ; b) $S_{\infty} = \frac{14}{3}$ ; c) $n = 64$

Show working and answer

a) 1. The first term is obtained with $n = 1$ : $u_{1} = \frac{2}{3} \cdot \frac{7}{8} = \frac{14}{24} = \frac{7}{12}$ .

b) 1. Each term is obtained from the previous one by multiplying by $\frac{7}{8}$ , so the ratio is $r = \frac{7}{8}$ , with $∣ r ∣ < 1$ , and the series converges.

Sum to infinity: $S_{\infty} = \frac{u _{1}}{1 - r} = \frac{\frac{7}{12}}{1 - \frac{7}{8}} = \frac{\frac{7}{12}}{\frac{1}{8}} = \frac{56}{12} = \frac{14}{3}$ .

c) 1. As $S_{n} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ , we have $S_{\infty} - S_{n} = \frac{u _{1} r ^{n}}{1 - r} = \frac{14}{3} (\frac{7}{8})^{n}$ .

We require $\frac{14}{3} (\frac{7}{8})^{n} < 0.001$ , that is, $(\frac{7}{8})^{n} < \frac{0.003}{14} \approx 0.000214286$ .
Taking logarithms; as $ln \frac{7}{8} < 0$ , the inequality reverses: $n > \frac{ln ( 0.000214286 )}{ln ( 7/8 )} = \frac{- 8.448}{- 0.13353} \approx 63.27$ .
The least integer is $n = 64$ .

Answer: a) $u_{1} = \frac{7}{12}$ ; b) $S_{\infty} = \frac{14}{3}$ ; c) $n = 64$

Exercise 62 HardIB · P2 TZ2 May 2010

A circle of radius 2 cm is divided into an infinite number of circular sectors. The areas of these sectors form a geometric sequence with common ratio $k$ . The angle of the first sector is $θ$ radians.

a) Verify that $θ = 2 π (1 - k)$ .

b) The perimeter of the third sector is half the perimeter of the first sector. Find the values of $k$ and $θ$ .

Show answer

Answer: $k \approx 0.456$ , $θ \approx 3.42$

Show working and answer

The area of the circle is $π r^{2} = 4 π$ . The sectors cover it completely, so the sum of their areas (an infinite geometric sequence) equals $4 π$ .
The area of a sector of angle $θ$ is $\frac{1}{2} r^{2} θ = \frac{1}{2} \cdot 4 \cdot θ = 2 θ$ . Hence the first term is $u_{1} = 2 θ$ .
Sum to infinity: $S_{\infty} = \frac{u _{1}}{1 - k}$ , with $∣ k ∣ < 1$ . Then $\frac{2 θ}{1 - k} = 4 π \Rightarrow θ = 2 π (1 - k) .$

As required.

The perimeter of a sector of angle $α$ is $2 r + r α = 4 + 2 α$ (two radii plus the arc).
Since the area of sector $n$ is $2 α_{n}$ , the angles also form a geometric sequence with ratio $k$ : $α_{n} = θ k^{n - 1}$ . The third angle is $θ k^{2}$ .
Condition: $4 + 2 θ k^{2} = \frac{1}{2} (4 + 2 θ) = 2 + θ$ , i.e. $θ (1 - 2 k^{2}) = 2$ .
Substitute $θ = 2 π (1 - k)$ : $2 π (1 - k) (1 - 2 k^{2}) = 2 \Rightarrow π (1 - k) (1 - 2 k^{2}) = 1.$
Solving the equation with the GDC (with $0 < k < 1$ , since the areas are positive): $k \approx 0.4557$ . There is no other solution in that interval (for $k > 1/ 2$ the product is negative, and for $k < 0.4557$ it is greater than 1).
Then $θ = 2 π (1 - 0.4557) \approx 3.42$ rad.

Answer: $k \approx 0.456$ , $θ \approx 3.42$

Exercise 63 HardIB · P1 TZ0 November 2010

The sum, $S_{n}$ , of the first $n$ terms of a geometric sequence, whose $n$ th term is $u_{n}$ , is given by $S_{n} = \frac{7 ^{n} - a ^{n}}{7 ^{n}}$ , where $a > 0$ .

a) Find an expression for $u_{n}$ .

b) Find the first term and the common ratio of the sequence.

c) Consider the sum to infinity of the sequence.

(i) Determine the values of $a$ for which the sum to infinity of the sequence exists.

(ii) Find the sum to infinity, when this sum exists.

Show answer

Answer: $u_{n} = (1 - \frac{a}{7}) (\frac{a}{7})^{n - 1}$

Answer: first term $u_{1} = \frac{7 - a}{7}$ , common ratio $r = \frac{a}{7}$

Answer: (i) $0 < a < 7$ ; (ii) $S_{\infty} = 1$

Show working and answer

We write $S_{n} = \frac{7 ^{n} - a ^{n}}{7 ^{n}} = 1 - (\frac{a}{7})^{n}$ .
For $n \geq 2$ , $u_{n} = S_{n} - S_{n - 1} = (\frac{a}{7})^{n - 1} - (\frac{a}{7})^{n} = (\frac{a}{7})^{n - 1} (1 - \frac{a}{7})$ .
For $n = 1$ : $u_{1} = S_{1} = 1 - \frac{a}{7}$ , which agrees with the formula above.

Answer: $u_{n} = (1 - \frac{a}{7}) (\frac{a}{7})^{n - 1}$

With $n = 1$ : $u_{1} = 1 - \frac{a}{7} = \frac{7 - a}{7}$ .
The common ratio is the repeated factor: $r = \frac{a}{7}$ .

Answer: first term $u_{1} = \frac{7 - a}{7}$ , common ratio $r = \frac{a}{7}$

(i) The sum to infinity exists if $∣ r ∣ < 1$ , i.e. $\frac{a}{7} < 1$ . Since $a > 0$ , this gives $0 < a < 7$ .

(ii) If $0 < a < 7$ , $(\frac{a}{7})^{n} \to 0$ as $n \to \infty$ , so $S_{\infty} = 1 - 0 = 1$ . (Using the formula: $\frac{u _{1}}{1 - r} = \frac{1 - a /7}{1 - a /7} = 1$ .)

Answer: (i) $0 < a < 7$ ; (ii) $S_{\infty} = 1$

Arithmetic and geometric sequences combined

Exercise 64 MediumIB · P1 TZ2 May 2005

The sum of the first $n$ terms of an arithmetic sequence ${u_{n}}$ is given by $S_{n} = 4 n^{2} - 2 n$ . Three terms of this sequence, $u_{2}$ , $u_{m}$ and $u_{32}$ , are consecutive terms of a geometric sequence. Find $m$ .

Show answer

Answer: $m = 7$

Show working and answer

For $n \geq 2$ , the $n$ th term is the difference of two consecutive sums: $u_{n} = S_{n} - S_{n - 1}$ .

$u_{n} = 4 n^{2} - 2 n - [4 (n - 1)^{2} - 2 (n - 1)] = 4 n^{2} - 2 n - 4 n^{2} + 8 n - 4 + 2 n - 2 = 8 n - 6$ .
For $n = 1$ : $u_{1} = S_{1} = 4 - 2 = 2$ , and the formula $8 n - 6$ also gives $2$ . Therefore $u_{n} = 8 n - 6$ for all $n$ (arithmetic sequence with $u_{1} = 2$ , $d = 8$ ).
The three terms: $u_{2} = 10$ and $u_{32} = 8 \cdot 32 - 6 = 250$ .
If $u_{2}, u_{m}, u_{32}$ are consecutive terms of a geometric sequence, the ratio between consecutive terms is constant: $\frac{u _{m}}{u _{2}} = \frac{u _{32}}{u _{m}} \Rightarrow u_{m}^{2} = u_{2} u_{32} = 10 \cdot 250 = 2500 \Rightarrow u_{m} = \pm 50.$
Since $u_{n} = 8 n - 6 > 0$ for $n \geq 1$ , we reject $u_{m} = - 50$ (it would give $m = - 44/8 < 0$ ). So $u_{m} = 50$ (ratio $5$ : $10, 50, 250$ ).
$8 m - 6 = 50 \Rightarrow 8 m = 56 \Rightarrow m = 7$ .

Answer: $m = 7$

Exercise 65 MediumIB · P1 TZ0 November 2009

a) The sum of the first six terms of an arithmetic sequence is equal to 81. The sum of the first eleven terms is equal to 231. Find the first term and the common difference.

b) The sum of the first two terms of a geometric sequence is equal to 1, while the sum of the first four terms is equal to 5. Given that all its terms are positive, find the first term and the common ratio.

c) The $r$ th term of a new sequence is defined as the product of the $r$ th term of the arithmetic sequence and the $r$ th term of the geometric sequence mentioned above. Verify that the $r$ th term of this new sequence is $(r + 1) \cdot 2^{r - 1}$ .

View solution

$S_{6} = \frac{6}{2} (2 u_{1} + 5 d) = 81 \Rightarrow 2 u_{1} + 5 d = 27$ .
$S_{11} = \frac{11}{2} (2 u_{1} + 10 d) = 11 (u_{1} + 5 d) = 231 \Rightarrow u_{1} + 5 d = 21$ .
Subtract the second equation from the first: $u_{1} = 27 - 21 = 6$ . Then $5 d = 15 \Rightarrow d = 3$ .

Answer a): $u_{1} = 6, d = 3$

Let $a$ be the first term and $r$ the common ratio. $S_{2} = a (1 + r) = 1$ and $S_{4} = a (1 + r + r^{2} + r^{3}) = a (1 + r) (1 + r^{2}) = 5$ .
Divide: $1 + r^{2} = 5 \Rightarrow r^{2} = 4 \Rightarrow r = \pm 2$ . Since all terms are positive, $r = 2$ (we reject $r = - 2$ , which would alternate in sign).
$a (1 + 2) = 1 \Rightarrow a = \frac{1}{3}$ .

Answer b): $a = \frac{1}{3}, r = 2$

The $r$ th term of the arithmetic sequence: $6 + 3 (r - 1) = 3 r + 3 = 3 (r + 1)$ .
The $r$ th term of the geometric sequence: $\frac{1}{3} \cdot 2^{r - 1}$ .
Their product: $3 (r + 1) \cdot \frac{1}{3} \cdot 2^{r - 1} = (r + 1) \cdot 2^{r - 1}$ .

As required.

Exercise 66 MediumIB · P1 TZ2 May 2011Proof

In an arithmetic sequence, the first term is $a$ and the common difference is $d$ , $d \neq = 0$ . The 3rd, 4th and 7th terms of this arithmetic sequence are the first three terms of a geometric sequence.

a) Verify that $a = - \frac{3}{2} d$ .

b) Verify that the 4th term of the geometric sequence is the 16th term of the arithmetic sequence.

View solution

Terms of the arithmetic sequence: $u_{3} = a + 2 d$ , $u_{4} = a + 3 d$ , $u_{7} = a + 6 d$ .

Three consecutive terms of a geometric sequence satisfy $(second)^{2} = (first) (third)$ : $(a + 3 d)^{2} = (a + 2 d) (a + 6 d) .$
Expand: $a^{2} + 6 a d + 9 d^{2} = a^{2} + 8 a d + 12 d^{2}$ .
Simplifying: $- 2 a d = 3 d^{2}$ . Since $d \neq = 0$ , we divide by $d$ : $- 2 a = 3 d$ , i.e. $a = - \frac{3}{2} d$ .

As required.

With $a = - \frac{3}{2} d$ : $u_{3} = \frac{1}{2} d$ , $u_{4} = \frac{3}{2} d$ , $u_{7} = \frac{9}{2} d$ .
The common ratio of the geometric sequence is $r = \frac{3 d /2}{d /2} = 3$ (check: $\frac{9 d /2}{3 d /2} = 3$ ).
The 4th term of the geometric sequence is $\frac{9}{2} d \cdot 3 = \frac{27}{2} d$ .
We seek $n$ with $u_{n} = a + (n - 1) d = \frac{27}{2} d$ : $- \frac{3}{2} d + (n - 1) d = \frac{27}{2} d \Rightarrow n - 1 = \frac{27}{2} + \frac{3}{2} = 15 \Rightarrow n = 16.$

As required.

Exercise 67 MediumIB · P2 TZ2 May 2013

In the arithmetic sequence ${u_{n} : n \in Z}$ , the first term is $u_{1} = 1.6$ and the common difference is $d = 1.5$ . In the geometric sequence ${v_{n} : n \in Z}$ , the first term is $v_{1} = 3$ and the common ratio is $r = 1.2$ .

a) Find an expression for $u_{n} - v_{n}$ in terms of $n$ .

b) Determine the set of values of $n$ for which $u_{n} > v_{n}$ .

c) Determine the greatest value of $u_{n} - v_{n}$ . Give your answer correct to four significant figures.

Show answer

Answer: $u_{n} - v_{n} = 1.5 n + 0.1 - 3 \cdot 1. 2^{n - 1}$

Answer: ${3, 4, 5, 6, 7, 8, 9}$

Answer: $1.642$

Show working and answer

Arithmetic sequence: $u_{n} = 1.6 + (n - 1) \cdot 1.5 = 1.5 n + 0.1$ .
Geometric sequence: $v_{n} = 3 \cdot 1. 2^{n - 1}$ .
Difference: $u_{n} - v_{n} = 1.5 n + 0.1 - 3 \cdot 1. 2^{n - 1}$ .

Answer: $u_{n} - v_{n} = 1.5 n + 0.1 - 3 \cdot 1. 2^{n - 1}$

We study the sign of $u_{n} - v_{n}$ using a table of values (GDC):

$n$	1	2	3	4	5	6	7	8	9	10
$u_{n} - v_{n}$	$- 1.4$	$- 0.5$	$0.28$	$0.916$	$1.379$	$1.635$	$1.642$	$1.351$	$0.701$	$- 0.379$

For $n \geq 10$ the difference remains negative, because the exponential term grows faster and faster than the linear term (this can also be seen from the graph). It is positive for $n = 3, 4, \dots, 9$ .

Answer: ${3, 4, 5, 6, 7, 8, 9}$

From the table, the difference increases up to $n = 7$ and then decreases, so the maximum occurs at $n = 7$ .
$u_{7} - v_{7} = 10.6 - 3 \cdot 1. 2^{6} = 10.6 - 8.957952 = 1.642048$ .

Answer: $1.642$

Exercise 68 MediumIB · P1 TZ2 May 2004

The three terms $a$ , $1$ , $b$ are in arithmetic sequence. The three terms $1$ , $a$ , $b$ are in geometric sequence. Find the values of $a$ and $b$ , given that $a \neq = b$ .

Show answer

Answer: $a = - 2, b = 4$

Show working and answer

In an arithmetic sequence the middle term is the mean of the other two: $1 = \frac{a + b}{2}$ , that is, $a + b = 2$ .
In a geometric sequence the ratio between consecutive terms is constant: $\frac{a}{1} = \frac{b}{a}$ , that is, $b = a^{2}$ .
Substituting into the first equation: $a + a^{2} = 2$ , so $a^{2} + a - 2 = 0$ , which factorises as $(a + 2) (a - 1) = 0$ . Thus $a = - 2$ or $a = 1$ .
If $a = 1$ , then $b = 1$ , and we would have $a = b$ , which the statement excludes. This value is rejected.
If $a = - 2$ , then $b = a^{2} = 4$ . Check: $- 2, 1, 4$ is arithmetic (difference $3$ ) and $1, - 2, 4$ is geometric (ratio $- 2$ ).

Answer: $a = - 2, b = 4$

Exercise 69 MediumIB · P1 TZ1 May 2011

In a geometric sequence $u_{1}, u_{2}, u_{3}, \dots$ $u_{1} = 27$ and its infinite sum is $\frac{81}{2}$ .

a) Find the common ratio of the geometric sequence.

An arithmetic sequence $v_{1}, v_{2}, v_{3}, \dots, \dots$ is such that $v_{2} = u_{2}$ and $v_{4} = u_{4}$ .

b) Find the greatest value of $N$ such that $n = 1 \sum N v_{n} > 0$ .

Show answer

Answer: a) $r = \frac{1}{3}$ ; b) $N = 7$

Show working and answer

(a)

Infinite sum: $\frac{u _{1}}{1 - r} = \frac{27}{1 - r} = \frac{81}{2}$ .
$1 - r = \frac{54}{81} = \frac{2}{3} \Rightarrow r = \frac{1}{3}$ (which satisfies $∣ r ∣ < 1$ ).

(b)

Terms of the geometric sequence: $u_{2} = 27 \cdot \frac{1}{3} = 9$ and $u_{4} = 27 \cdot \frac{1}{27} = 1$ .
In the arithmetic sequence, $v_{2} = 9$ and $v_{4} = 1$ ; since $v_{4} = v_{2} + 2 d$ , $1 = 9 + 2 d \Rightarrow d = - 4$ .
$v_{1} = v_{2} - d = 13$ .
Sum: $n = 1 \sum N v_{n} = \frac{N}{2} [2 \cdot 13 + (N - 1) (- 4)] = \frac{N}{2} (30 - 4 N) = N (15 - 2 N)$ .
We impose $N (15 - 2 N) > 0$ . As $N > 0$ , we need $15 - 2 N > 0 \Rightarrow N < 7.5$ .
The greatest integer is $N = 7$ (check: $7 \cdot 1 = 7 > 0$ ; for $N = 8$ : $8 \cdot (- 1) = - 8 < 0$ ).

Answer: a) $r = \frac{1}{3}$ ; b) $N = 7$

Exercise 70 MediumIB · P1 TZ1 May 2017

An arithmetic sequence $u_{1}, u_{2}, u_{3}, \dots$ has $u_{1} = 1$ and common difference $d \neq = 0$ .

Given that $u_{2}$ , $u_{3}$ and $u_{6}$ are the first three terms of a geometric sequence.

a) Find the value of $d$ .

Given that $u_{N} = - 15$

b) Determine the value of $r = 1 \sum N u_{r}$ .

Show answer

Answer: a) $d = - 2$ ; b) $r = 1 \sum N u_{r} = - 63$

Show working and answer

a) 1. The terms are $u_{2} = 1 + d$ , $u_{3} = 1 + 2 d$ , $u_{6} = 1 + 5 d$ . 2. If they are in geometric sequence, the ratio of consecutive terms is constant: $\frac{u _{3}}{u _{2}} = \frac{u _{6}}{u _{3}}$ , that is, $u_{3}^{2} = u_{2} u_{6}$ . 3. $(1 + 2 d)^{2} = (1 + d) (1 + 5 d)$ gives $1 + 4 d + 4 d^{2} = 1 + 6 d + 5 d^{2}$ . 4. Simplifying: $d^{2} + 2 d = 0$ , so $d (d + 2) = 0$ . 5. Since $d \neq = 0$ , $d = - 2$ . (Check: the terms are $- 1, - 3, - 9$ , with ratio $3$ .)

b) 1. $u_{N} = 1 + (N - 1) (- 2) = - 15$ , so $N - 1 = 8$ and $N = 9$ . 2. $S_{9} = \frac{9}{2} (u_{1} + u_{9}) = \frac{9}{2} (1 - 15) = \frac{9}{2} \cdot (- 14) = - 63$ .

Answer: a) $d = - 2$ ; b) $r = 1 \sum N u_{r} = - 63$

Exercise 71 MediumIB · P1 TZ2 May 2017

The $1^{st}$ , $4^{th}$ and $8^{th}$ terms of an arithmetic sequence with common difference $d$ , $d \neq = 0$ , are the first three terms of a geometric sequence with common ratio $r$ .

Given that the $1^{st}$ term of both sequences is $9$ , find

a) the value of $d$

b) the value of $r$ .

Show answer

Answer: a) $d = 1$ ; b) $r = \frac{4}{3}$

Show working and answer

Let $u_{1} = 9$ be the first term of the arithmetic sequence; then $u_{4} = 9 + 3 d$ and $u_{8} = 9 + 7 d$ .

a) 1. The terms $9, 9 + 3 d, 9 + 7 d$ form a geometric sequence, so $(9 + 3 d)^{2} = 9 (9 + 7 d)$ . 2. Expanding: $81 + 54 d + 9 d^{2} = 81 + 63 d$ . 3. Simplifying: $9 d^{2} - 9 d = 0$ , so $9 d (d - 1) = 0$ . 4. Since $d \neq = 0$ , $d = 1$ .

b) 1. With $d = 1$ the three terms are $9, 12, 16$ . 2. The ratio is $r = \frac{12}{9} = \frac{4}{3}$ . Check: $\frac{16}{12} = \frac{4}{3}$ .

Answer: a) $d = 1$ ; b) $r = \frac{4}{3}$

Exercise 72 MediumIB · P1 May 1998

The first, second and $n$ th terms of an arithmetic sequence are $2$ , $6$ and $58$ respectively.

a) Calculate the value of $n$ .

b) For that value of $n$ , calculate the exact value of the sum of the $n$ terms of a geometric sequence whose first term is $2$ and common ratio is $\frac{1}{2}$ .

Show answer

Answer: a) $n = 15$ ; b) $S_{15} = \frac{32767}{8192}$

Show working and answer

First term $2$ and common difference $d = 6 - 2 = 4$ .
$u_{n} = 2 + (n - 1) \cdot 4 = 58 \Rightarrow n - 1 = 14 \Rightarrow n = 15$ .

Sum of a geometric sequence: $S_{n} = \frac{u _{1} ( 1 - r ^{n} )}{1 - r}$ with $u_{1} = 2$ , $r = \frac{1}{2}$ , $n = 15$ .
$S_{15} = \frac{2 ( 1 - \frac{1}{2 ^{15}} )}{\frac{1}{2}} = 4 (1 - \frac{1}{32768}) = 4 - \frac{1}{8192} = \frac{32767}{8192}$ .

Answer: a) $n = 15$ ; b) $S_{15} = \frac{32767}{8192}$

Exercise 73 HardIB · P2 November 1997

a) Verify that $p = 2$ is a solution of the equation $p^{3} + p^{2} - 5 p - 2 = 0$ .

b) Find the values of $a$ and $b$ for which the following holds: $p^{3} + p^{2} - 5 p - 2 = (p - 2) (p^{2} + a p + b)$ .

c) Hence, find the other two roots of the equation $p^{3} + p^{2} - 5 p - 2 = 0$ .

d) The common difference of an arithmetic sequence is equal to $p$ . The common ratio of a geometric sequence is also equal to $p$ . The first term of both sequences is 1.

(i) Write down, in terms of $p$ , the first four terms of each sequence.

(ii) If the sum of the third and fourth terms of the arithmetic sequence is equal to the sum of the third and fourth terms of the geometric sequence, find the three possible values of $p$ .

(iii) For which value of $p$ , from those obtained in part (d)(ii), does the sum to infinity of the geometric sequence exist?

(iv) For this same value of $p$ , find the sum of the first 20 terms of the arithmetic sequence, giving your answer in the form $a + b c$ , where $a, b, c \in Z$ .

View solution

a) We substitute $p = 2$ : $2^{3} + 2^{2} - 5 \cdot 2 - 2 = 8 + 4 - 10 - 2 = 0$ . The equation holds, so $p = 2$ is a solution.

b) We expand the right-hand side: $(p - 2) (p^{2} + a p + b) = p^{3} + a p^{2} + b p - 2 p^{2} - 2 a p - 2 b = p^{3} + (a - 2) p^{2} + (b - 2 a) p - 2 b .$ We equate coefficients with $p^{3} + p^{2} - 5 p - 2$ :

$p^{2}$ : $a - 2 = 1 \Rightarrow a = 3$ .
Constant term: $- 2 b = - 2 \Rightarrow b = 1$ .
Check on $p$ : $b - 2 a = 1 - 6 = - 5$ . Correct.

Answer b): $a = 3, b = 1$

c) The equation is $(p - 2) (p^{2} + 3 p + 1) = 0$ . The other roots satisfy $p^{2} + 3 p + 1 = 0$ : $p = \frac{- 3 \pm 9 - 4}{2} = \frac{- 3 \pm 5}{2} .$

Answer c): $p = \frac{- 3 + 5}{2}$ and $p = \frac{- 3 - 5}{2}$

d) (i) Arithmetic sequence ( $u_{1} = 1$ , $d = p$ ): $1, 1 + p, 1 + 2 p, 1 + 3 p$ . Geometric sequence ( $u_{1} = 1$ , $r = p$ ): $1, p, p^{2}, p^{3}$ .

(ii) We equate the sum of the third and fourth terms: $(1 + 2 p) + (1 + 3 p) = p^{2} + p^{3} \Rightarrow p^{3} + p^{2} - 5 p - 2 = 0.$ This is the equation from the earlier parts, so the three values are $p = 2$ , $p = \frac{- 3 + 5}{2}$ and $p = \frac{- 3 - 5}{2}$ .

(iii) The sum to infinity of the geometric sequence exists if $∣ p ∣ < 1$ .

$p = 2$ : $∣ p ∣ > 1$ , no.
$p = \frac{- 3 - 5}{2} \approx - 2.62$ : $∣ p ∣ > 1$ , no.
$p = \frac{- 3 + 5}{2} \approx - 0.382$ : $∣ p ∣ < 1$ , yes.

Answer (iii): $p = \frac{- 3 + 5}{2}$

(iv) With $u_{1} = 1$ and $d = p$ : $S_{20} = \frac{20}{2} [2 + 19 p] = 20 + 190 p$ . We substitute $p = \frac{- 3 + 5}{2}$ : $S_{20} = 20 + 95 (- 3 + 5) = 20 - 285 + 955 = - 265 + 955 .$

Answer (iv): $S_{20} = - 265 + 955$

Exercise 74 HardIB · P2 TZ0 November 2014

The seventh, third and first terms of an arithmetic sequence are the first three terms of a geometric sequence. In the arithmetic sequence, the first term is $a$ and the common difference, which is non-zero, is $d$ .

a) Show that $d = \frac{a}{2}$ . The seventh term of the arithmetic sequence is 3. The sum of the first $n$ terms of the arithmetic sequence exceeds the sum of the first $n$ terms of the geometric sequence by at least 200.

b) Find the least value of $n$ for which this occurs.

Show answer

Answer: $d = \frac{a}{2}$ ; $n = 32$

Show working and answer

a) 1. Terms of the arithmetic sequence: $u_{1} = a$ , $u_{3} = a + 2 d$ , $u_{7} = a + 6 d$ .

If $u_{7}, u_{3}, u_{1}$ are consecutive terms of a geometric sequence, the ratio between consecutive terms is constant: $\frac{a + 2 d}{a + 6 d} = \frac{a}{a + 2 d}$ , that is, $(a + 2 d)^{2} = a (a + 6 d)$ .
Expanding: $a^{2} + 4 a d + 4 d^{2} = a^{2} + 6 a d$ , so $4 d^{2} = 2 a d$ .
As $d \neq = 0$ , we divide by $2 d$ : $2 d = a$ , that is, $d = \frac{a}{2}$ .

b) 1. $u_{7} = a + 6 d = a + 3 a = 4 a = 3$ , so $a = \frac{3}{4}$ and $d = \frac{3}{8}$ .

Geometric sequence: first term $u_{7} = 3$ ; second term $u_{3} = a + 2 d = 2 a = \frac{3}{2}$ . Ratio $r = \frac{1}{2}$ .
Arithmetic sum: $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [\frac{3}{2} + \frac{3 ( n - 1 )}{8}] = \frac{3 n ( n + 3 )}{16}$ .
Geometric sum: $T_{n} = \frac{3 ( 1 - ( \frac{1}{2} ) ^{n} )}{1 - \frac{1}{2}} = 6 (1 - 2^{- n})$ .
Condition: $\frac{3 n ( n + 3 )}{16} - 6 (1 - 2^{- n}) \geq 200$ .
We test values: for $n = 31$ , $\frac{3 \cdot 31 \cdot 34}{16} = 197.4875$ and the difference is approximately $191.49 < 200$ . For $n = 32$ , $\frac{3 \cdot 32 \cdot 35}{16} = 210$ and the difference is $210 - 6 (1 - 2^{- 32}) \approx 204 > 200$ .
As the difference increases with $n$ , the least value is $n = 32$ .

Answer: $d = \frac{a}{2}$ ; $n = 32$

Exercise 75 HardIB · P2 TZ1 May 2005

The first three terms of a geometric sequence are the first, eleventh and sixteenth terms of an arithmetic sequence. All the terms of the geometric sequence are different. The infinite sum of the geometric sequence is $18$ .

a) Calculate the common ratio of the geometric sequence.

b) Calculate the common difference of the arithmetic sequence.

Show answer

Answer: a) $r = \frac{1}{2}$ ; b) $d = - \frac{9}{20} = - 0.45$

Show working and answer

a) 1. Let $a$ be the first term of the geometric sequence and $r$ its common ratio. Let $d$ be the common difference of the arithmetic sequence, whose first term is $a$ . Then the first, eleventh and sixteenth terms are $a, a + 10 d, a + 15 d$ . 2. These are the first three terms of the geometric sequence: $a r = a + 10 d$ and $a r^{2} = a + 15 d$ . 3. Therefore $a (r - 1) = 10 d$ and $a (r^{2} - 1) = 15 d$ . 4. As the terms are distinct, $a \neq = 0$ and $r \neq = 1$ (and then $d \neq = 0$ ). Dividing the second equation by the first: $\frac{a ( r - 1 ) ( r + 1 )}{a ( r - 1 )} = \frac{15 d}{10 d}$ , that is, $r + 1 = \frac{3}{2}$ . 5. Hence $r = \frac{1}{2}$ .

b) 1. The infinite sum is $\frac{a}{1 - r} = 18$ , that is, $\frac{a}{1/2} = 18$ , so $a = 9$ . 2. From $a (r - 1) = 10 d$ : $9 (- \frac{1}{2}) = 10 d$ , that is, $d = - \frac{9}{20}$ . 3. Check: $a + 10 d = 9 - 4.5 = 4.5 = a r$ and $a + 15 d = 9 - 6.75 = 2.25 = a r^{2}$ .

Answer: a) $r = \frac{1}{2}$ ; b) $d = - \frac{9}{20} = - 0.45$

Exercise 76 HardIB · P1 TZ2 May 2015

The cubic equation $x^{3} + p x^{2} + q x + c = 0$ has roots $α$ , $β$ , $γ$ . By expanding $(x - α) (x - β) (x - γ)$ , show that:

a) (i) $p = - (α + β + γ)$

(ii) $q = α β + β γ + γ α$

(iii) $c = - α β γ$ Now it is known that $p = - 6$ and $q = 18$ for parts (b) and (c).

b) (i) In the case where the three roots $α$ , $β$ , $γ$ form an arithmetic sequence, show that one of the roots is $2$ .

(ii) Hence, find the value of $c$ .

c) In the other case, the three roots $α$ , $β$ , $γ$ form a geometric sequence. Find the value of $c$ .

Show answer

Answer: a) proved; b) (i) one root is $2$ , (ii) $c = - 20$ ; c) $c = - 27$

Show working and answer

a) We expand step by step:

$(x - α) (x - β) = x^{2} - (α + β) x + α β$ .
Multiplying by $(x - γ)$ : $x^{3} - (α + β) x^{2} + α β x - γ x^{2} + γ (α + β) x - α β γ$ $= x^{3} - (α + β + γ) x^{2} + (α β + β γ + γ α) x - α β γ .$
Comparing with $x^{3} + p x^{2} + q x + c$ (the same monic polynomial with those roots): (i) $p = - (α + β + γ)$ ; (ii) $q = α β + β γ + γ α$ ; (iii) $c = - α β γ$ .

b) (i) With $p = - 6$ and the roots in arithmetic sequence:

We write the roots as $a - d, a, a + d$ .
By (i): $α + β + γ = - p = 6 \Rightarrow 3 a = 6 \Rightarrow a = 2$ . So one of the roots is $2$ .

(ii) As $2$ is a root, we substitute into $x^{3} - 6 x^{2} + 18 x + c = 0$ :

$8 - 24 + 36 + c = 0 \Rightarrow c = - 20$ .

(Check with the roots $2 - d, 2, 2 + d$ : $q = 12 - d^{2} = 18 \Rightarrow d^{2} = - 6$ , and $c = - 2 (4 - d^{2}) = - 2 \cdot 10 = - 20$ .)

c) Roots in geometric sequence: $\frac{a}{r}, a, a r$ .

Product: $α β γ = a^{3}$ , and by (iii) $c = - a^{3}$ .
Sum: $a (\frac{1}{r} + 1 + r) = - p = 6$ .
Sum of the pairwise products: $\frac{a ^{2}}{r} + a^{2} r + a^{2} = a^{2} (\frac{1}{r} + 1 + r) = q = 18$ .
Dividing (3) by (2): $a = \frac{18}{6} = 3$ .
Then $c = - a^{3} = - 27$ .

Answer: a) proved; b) (i) one root is $2$ , (ii) $c = - 20$ ; c) $c = - 27$

Exercise 77 HardIB · P1 TZ2 May 2018

The geometric sequence $u_{1}, u_{2}, u_{3}, \dots$ has common ratio $r$ .

Consider the sequence $A = {a_{n} = lo g_{2} ∣ u_{n} ∣ : n \in Z^{+}}$ .

a) Show that $A$ is an arithmetic sequence, and state its common difference, $d$ , in terms of $r$ .

In a particular geometric sequence, $u_{1} = 3$ and the sum of the infinitely many terms is equal to $4$ .

b) Find the value of $d$ .

Show answer

Answer: a) $d = lo g_{2} ∣ r ∣$ ; b) $d = - 2$

Show working and answer

a) 1. The terms are $u_{n} = u_{1} r^{n - 1}$ , so $a_{n} = lo g_{2} ∣ u_{n} ∣ = lo g_{2} (∣ u_{1} ∣ ∣ r ∣^{n - 1})$ . 2. By the properties of logarithms: $a_{n} = lo g_{2} ∣ u_{1} ∣ + (n - 1) lo g_{2} ∣ r ∣$ . 3. This has the form $a_{n} = a_{1} + (n - 1) d$ with $a_{1} = lo g_{2} ∣ u_{1} ∣$ . Equivalently, $a_{n + 1} - a_{n} = lo g_{2} \frac{∣ u _{n + 1} ∣}{∣ u _{n} ∣} = lo g_{2} ∣ r ∣$ , which is constant and independent of $n$ . 4. Hence $A$ is an arithmetic sequence with common difference $d = lo g_{2} ∣ r ∣$ .

Hence proved. $d = lo g_{2} ∣ r ∣$ .

b) 1. The sum of infinitely many terms of a geometric sequence (with $∣ r ∣ < 1$ ) is $S_{\infty} = \frac{u _{1}}{1 - r}$ . So $\frac{3}{1 - r} = 4$ . 2. Solving: $3 = 4 - 4 r$ , so $r = \frac{1}{4}$ (which satisfies $∣ r ∣ < 1$ ). 3. $d = lo g_{2} \frac{1}{4} = lo g_{2} 2^{- 2} = - 2$ .

Answer: a) $d = lo g_{2} ∣ r ∣$ ; b) $d = - 2$

Exercise 78 HardIB · P1 TZ0 November 2015

A polynomial function is given by $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{n} x^{n}$ .

The roots of the polynomial equation $f (x) = 0$ are consecutive terms of a geometric sequence with common ratio equal to $\frac{1}{2}$ and first term $2$ . Given that $a_{n - 1} = - 63$ and $a_{n} = 16$ , find

a) the degree of the polynomial;

b) the value of $a_{0}$ .

Show answer

Answer: a) $n = 6$ ; b) $a_{0} = \frac{1}{32}$

Show working and answer

The roots are $2, 1, \frac{1}{2}, \dots$ , that is, $2 (\frac{1}{2})^{k}$ with $k = 0, 1, \dots, n - 1$ (a polynomial of degree $n$ has $n$ roots).
By Vieta's formulas, the sum of the roots is $- \frac{a _{n - 1}}{a _{n}} = - \frac{- 63}{16} = \frac{63}{16}$ .
Sum of the first $n$ terms of the geometric sequence: $S_{n} = \frac{2 ( 1 - ( \frac{1}{2} ) ^{n} )}{1 - \frac{1}{2}} = 4 (1 - \frac{1}{2 ^{n}})$ .
We equate: $4 (1 - \frac{1}{2 ^{n}}) = \frac{63}{16} \Rightarrow 1 - \frac{1}{2 ^{n}} = \frac{63}{64} \Rightarrow \frac{1}{2 ^{n}} = \frac{1}{64}$ , so $2^{n} = 64$ and $n = 6$ .

The product of the roots is $(- 1)^{n} \frac{a _{0}}{a _{n}}$ . With $n = 6$ this is $\frac{a _{0}}{16}$ .
Product of the roots: $2^{6} \cdot (\frac{1}{2})^{0 + 1 + 2 + 3 + 4 + 5} = 2^{6} \cdot 2^{- 15} = 2^{- 9}$ .
Therefore $\frac{a _{0}}{16} = 2^{- 9} \Rightarrow a_{0} = 2^{4} \cdot 2^{- 9} = 2^{- 5} = \frac{1}{32}$ .

Answer: a) $n = 6$ ; b) $a_{0} = \frac{1}{32}$

Exercise 79 HardIB · P1 TZ1 May 2015

Let ${u_{n}}$ , $n \in Z^{+}$ , be an arithmetic sequence with first term $a$ and common difference $d$ , where $d \neq = 0$ . Let another sequence ${v_{n}}$ , $n \in Z^{+}$ , be defined by $v_{n} = 2^{u_{n}}$ .

a) (i) Show that $\frac{v _{n + 1}}{v _{n}}$ is a constant.

(ii) Write down the first term of the sequence ${v_{n}}$ .

(iii) Write down a formula for $v_{n}$ in terms of $a$ , $d$ and $n$ . Let $S_{n}$ be the sum of the first $n$ terms of the sequence ${v_{n}}$ .

b) (i) Write $S_{n}$ in terms of $a$ , $d$ and $n$ .

(ii) Find the values of $d$ for which $i = 1 \sum \infty v_{i}$ exists. It is now known that $i = 1 \sum \infty v_{i}$ exists and is denoted by $S_{\infty}$ .

(iii) Write $S_{\infty}$ in terms of $a$ and $d$ .

(iv) Given that $S_{\infty} = 2^{a + 1}$ , calculate the value of $d$ . Let ${w_{n}}$ , $n \in Z^{+}$ , be a geometric sequence with first term $p$ and common ratio $q$ , where $p$ and $q$ are both greater than zero. Now let another sequence ${z_{n}}$ be defined by $z_{n} = ln w_{n}$ .

c) Calculate $i = 1 \sum n z_{i}$ , giving your answer in the form $ln k$ , where $k$ is in terms of $n$ , $p$ and $q$ .

Show answer

Answer: a) (i) $2^{d}$ ; (ii) $2^{a}$ ; (iii) $v_{n} = 2^{a + (n - 1) d}$ ; b) (i) $S_{n} = \frac{2 ^{a} ( 2 ^{n d} - 1 )}{2 ^{d} - 1}$ ; (ii) $d < 0$ ; (iii) $S_{\infty} = \frac{2 ^{a}}{1 - 2 ^{d}}$ ; (iv) $d = - 1$ ; c) $ln (p^{n} q^{\frac{n ( n - 1 )}{2}})$

Show working and answer

a) (i) $u_{n} = a + (n - 1) d$ , so $v_{n} = 2^{u_{n}}$ .

$\frac{v _{n + 1}}{v _{n}} = \frac{2 ^{a + n d}}{2 ^{a + (n - 1) d}} = 2^{n d - (n - 1) d} = 2^{d}$ , which is constant (it does not depend on $n$ ).

(ii) $v_{1} = 2^{u_{1}} = 2^{a}$ .

(iii) It is a geometric sequence with first term $2^{a}$ and ratio $2^{d}$ : $v_{n} = 2^{a} (2^{d})^{n - 1} = 2^{a + (n - 1) d}$ .

b) (i) Sum of a geometric sequence $S_{n} = \frac{v _{1} ( r ^{n} - 1 )}{r - 1}$ :

$S_{n} = \frac{2 ^{a} ( 2 ^{n d} - 1 )}{2 ^{d} - 1}$ .

(ii) The sum to infinity exists if $∣ r ∣ < 1$ , that is $∣ 2^{d} ∣ < 1$ . As $2^{d} > 0$ , it is enough that $2^{d} < 1$ , i.e. $d < 0$ .

(iii) $S_{\infty} = \frac{v _{1}}{1 - r} = \frac{2 ^{a}}{1 - 2 ^{d}}$ .

(iv) $\frac{2 ^{a}}{1 - 2 ^{d}} = 2^{a + 1} = 2 \cdot 2^{a}$ .

Dividing by $2^{a}$ : $\frac{1}{1 - 2 ^{d}} = 2 \Rightarrow 1 - 2^{d} = \frac{1}{2} \Rightarrow 2^{d} = \frac{1}{2}$ .
Therefore $d = - 1$ , which satisfies $d < 0$ .

c) $w_{n} = p q^{n - 1}$ , so $z_{i} = ln w_{i} = ln p + (i - 1) ln q$ (properties of the logarithm), an arithmetic sequence with first term $ln p$ and difference $ln q$ .

$i = 1 \sum n z_{i} = n ln p + ln q i = 1 \sum n (i - 1) = n ln p + \frac{n ( n - 1 )}{2} ln q$ .
Using $m ln A = ln A^{m}$ and $ln A + ln B = ln (A B)$ : $ln (p^{n} q^{\frac{n ( n - 1 )}{2}})$ .

Exercise 80 HardIB · P1 TZ1 May 2022

Consider the series $ln x + p ln x + \frac{1}{3} ln x + \dots$ where $x \in R$ , $x > 1$ and $p \in R$ , $p \neq = 0$ .

a) Consider the case where the series is geometric.

(i) Show that $p = \pm \frac{1}{3}$ .

(ii) Given that $p > 0$ and $S_{\infty} = 3 + 3$ , calculate the value of $x$ .

b) Now consider the case where the series is arithmetic with common difference $d$ .

(i) Show that $p = \frac{2}{3}$ .

(ii) Write $d$ in the form $k ln x$ , where $k \in Q$ .

(iii) The sum of the first $n$ terms of the series is $- 3 ln x$ . Calculate the value of $n$ .

Show answer

Answer: a) (ii) $x = e^{2}$ ; b) (ii) $d = - \frac{1}{3} ln x$ ; (iii) $n = 9$

Show working and answer

a) (i)

In a geometric sequence the ratio of consecutive terms is constant: $\frac{p ln x}{ln x} = \frac{\frac{1}{3} ln x}{p ln x}$ (with $ln x \neq = 0$ since $x > 1$ ).
So $p = \frac{1}{3 p}$ , hence $p^{2} = \frac{1}{3}$ and $p = \pm \frac{1}{3}$ .

Q.E.D.

(ii)

With $p > 0$ , the ratio is $r = p = \frac{1}{3}$ and $u_{1} = ln x$ . As $∣ r ∣ < 1$ , $S_{\infty} = \frac{u _{1}}{1 - r} = \frac{ln x}{1 - \frac{1}{3}}$ .
We equate: $\frac{ln x}{1 - \frac{1}{3}} = 3 + 3$ , so $ln x = (3 + 3) (1 - \frac{1}{3}) = 3 - 3 + 3 - 1 = 2$ .
Hence $x = e^{2}$ .

b) (i)

In an arithmetic sequence the difference is constant: $p ln x - ln x = \frac{1}{3} ln x - p ln x$ .
Dividing by $ln x \neq = 0$ : $p - 1 = \frac{1}{3} - p$ , so $2 p = \frac{4}{3}$ and $p = \frac{2}{3}$ .

Q.E.D.

(ii) $d = p ln x - ln x = (\frac{2}{3} - 1) ln x = - \frac{1}{3} ln x$ , that is, $k = - \frac{1}{3}$ .

(iii)

Arithmetic sum formula: $S_{n} = \frac{n}{2} [2 u_{1} + (n - 1) d] = \frac{n}{2} [2 ln x - \frac{n - 1}{3} ln x]$ .
Equating to $- 3 ln x$ and dividing by $ln x > 0$ : $\frac{n}{2} (2 - \frac{n - 1}{3}) = - 3$ , that is, $\frac{n ( 7 - n )}{6} = - 3$ .
We get $n (7 - n) = - 18$ , i.e. $n^{2} - 7 n - 18 = 0$ , which factorises as $(n - 9) (n + 2) = 0$ .
As $n$ is positive, $n = 9$ (the solution $n = - 2$ is discarded).

Answer: a) (ii) $x = e^{2}$ ; b) (ii) $d = - \frac{1}{3} ln x$ ; (iii) $n = 9$

Compound interest, growth and depreciation

Exercise 81 MediumIB · P1 TZ0 November 2006

An amount of $100$ dollars is invested.

a) If interest is compounded annually at an annual rate of $5 %$ , find the total value $V$ of the investment after $20$ years.

b) If interest is compounded monthly at a monthly rate of $\frac{5}{12} %$ , find the minimum number of months after which the value of the investment exceeds $V$ .

Show answer

Answer: a) $V \approx 265.33$ dollars; b) $235$ months.

Show working and answer

a) Annual compound interest: $V = 100 \cdot 1.0 5^{20}$ .

$V \approx 265.33$ dollars.

b) With monthly interest of $\frac{5}{12} %$ , the value after $n$ months is $100 (1 + \frac{0.05}{12})^{n}$ .

We require $100 (1 + \frac{0.05}{12})^{n} > 265.33$ , that is, $(1.0041 6)^{n} > 2.6533$ .
Taking logarithms: $n ln 1.0041 6 > ln 2.6533 \Rightarrow n > \frac{0.97580}{0.0041580} \approx 234.7$ .
As $n$ is an integer: $n = 235$ . Check: with $234$ months the value is less than $V$ , and with $235$ it exceeds it.

Answer: a) $V \approx 265.33$ dollars; b) $235$ months.

Exercise 82 MediumIB · P2 TZ1 May 2022

In this question, give all answers correct to two decimal places. Sam invests $1700$ dollars in a savings account that pays a nominal annual interest rate of $2.74 %$ compounded semi-annually. Sam makes no further payments or withdrawals from this account.

a) Find the amount Sam will have in the account after $10$ years. David also invests $1700$ dollars in a savings account that pays a nominal annual interest rate of $r %$ , compounded annually. David makes no further payments or withdrawals from this account.

b) Find the value of $r$ required for the amount in David's account after $10$ years to equal the amount in Sam's account.

c) Calculate the interest David will earn over the $10$ years.

Show answer

Answer: a) $2231.71$ dollars; b) $r = 2.76 %$ ; c) $531.71$ dollars

Show working and answer

a) 1. Compound interest semi-annually: $2$ periods per year, with a rate of $\frac{2.74}{2} = 1.37 %$ per period, and $10 \cdot 2 = 20$ periods.

Final amount: $A = 1700 (1 + \frac{2.74}{200})^{20} = 1700 \cdot (1.0137)^{20} \approx 2231.71$ .

b) 1. With annual interest, David's amount is $1700 (1 + \frac{r}{100})^{10}$ . We equate it to Sam's: $1700 (1 + \frac{r}{100})^{10} = 2231.71$ .

Simplifying: $(1 + \frac{r}{100})^{10} = (1.0137)^{20} \approx 1.31277$ , so $1 + \frac{r}{100} = 1.3127 7^{1/10} \approx 1.027588$ .
Hence $r \approx 2.76$ .

c) 1. David has the same final amount as Sam, so the interest is $2231.71 - 1700 = 531.71$ .

Answer: a) $2231.71$ dollars; b) $r = 2.76 %$ ; c) $531.71$ dollars

Exercise 83 HardIB · P2 TZ1 May 2018

The number of fish in a certain lake is known to decrease by $7%$ each year, unless some new fish are added. At the end of each year, $250$ new fish are added to the lake. At the start of 2018, there are $2500$ fish in the lake.

a) Show that there will be approximately $2645$ fish in the lake at the start of 2020.

b) Find the approximate number of fish in the lake at the start of 2042.

Show answer

Answer: a) $\approx 2645$ fish; b) $\approx 3384$ fish

Show working and answer

Let $P_{n}$ be the number of fish at the start of the year $2018 + n$ . Each year it is multiplied by $0.93$ (a decrease of $7%$ ), and at the end $250$ fish are added: $P_{n + 1} = 0.93 P_{n} + 250$ , with $P_{0} = 2500$ .

a) 1. At the start of 2019: $P_{1} = 0.93 \cdot 2500 + 250 = 2325 + 250 = 2575$ . 2. At the start of 2020: $P_{2} = 0.93 \cdot 2575 + 250 = 2394.75 + 250 = 2644.75 \approx 2645$ .

Q.E.D.

b) 1. Expanding the recurrence: $P_{n} = 2500 \cdot 0.9 3^{n} + 250 (1 + 0.93 + \dots + 0.9 3^{n - 1})$ . 2. The sum in brackets is geometric with ratio $0.93$ : $\frac{1 - 0.9 3 ^{n}}{1 - 0.93} = \frac{1 - 0.9 3 ^{n}}{0.07}$ . 3. The start of 2042 corresponds to $n = 24$ , with $0.9 3^{24} \approx 0.17522$ . 4. $P_{24} = 2500 \cdot 0.17522 + 250 \cdot \frac{1 - 0.17522}{0.07} \approx 438.05 + 2945.6 \approx 3383.7$ . 5. Rounding to the nearest integer, there are about $3384$ fish.

Answer: a) $\approx 2645$ fish; b) $\approx 3384$ fish

Exercise 84 HardIB · P2 TZ0 November 2017

Phil takes out a bank loan of $150000$ dollars to buy a house, at an annual interest rate of $3.5 %$ . Interest is calculated at the end of each year and added to the outstanding amount.

a) Find the amount Phil pays the bank after $20$ years. Give your answer to the nearest dollar. To repay the loan, Phil makes annual deposits of $P$ dollars at the end of each year into a savings account, which pays an annual interest rate of $2 %$ . He makes his first deposit at the end of the first year after taking out the loan.

b) Show that the total value of Phil's savings after $20$ years is $\frac{( 1.0 2 ^{20} - 1 ) P}{( 1.02 - 1 )}$ .

c) Given that Phil's aim is to own the house after $20$ years, calculate the value of $P$ to the nearest dollar. David visits a different bank and obtains a deposit of $Q$ dollars, at an annual interest rate of $2.8 %$ .

d) (i) David wishes to withdraw $5000$ dollars at the end of each year for a period of $n$ years. Show that an expression for the minimum value of $Q$ is $\frac{5000}{1.028} + \frac{5000}{1.02 8 ^{2}} + \dots + \frac{5000}{1.02 8 ^{n}}$ .

(ii) Using the above or otherwise, calculate the minimum value of $Q$ that would allow David to withdraw $5000$ dollars annually indefinitely. Give your answer to the nearest dollar.

Show answer

Answer: a) $298468$ dollars; b) proved; c) $P = 12284$ dollars; d) (i) proved; (ii) $Q = 178571$ dollars.

Show working and answer

a) This is annual compound interest: the debt is multiplied by $1.035$ each year.

$150000 \cdot 1.03 5^{20} \approx 298468.3$ .
To the nearest dollar: $298468$ dollars.

b) Each deposit $P$ earns compound interest of $2 %$ until year 20.

The deposit at the end of year 1 remains for 19 years, the one at the end of year 2 for 18 years, ..., and the last one (at the end of year 20) earns no interest.
Total value: $P \cdot 1.0 2^{19} + P \cdot 1.0 2^{18} + \dots + P \cdot 1.02 + P$ , a geometric series with first term $P$ , ratio $1.02$ and $20$ terms.
Using $S_{n} = \frac{u _{1} ( r ^{n} - 1 )}{r - 1}$ : $\frac{P ( 1.0 2 ^{20} - 1 )}{1.02 - 1}$ .

c) We equate the savings to the debt:

$\frac{P ( 1.0 2 ^{20} - 1 )}{0.02} = 298468.3$ .
$P = \frac{298468.3 \cdot 0.02}{1.0 2 ^{20} - 1} = \frac{5969.37}{0.485947} \approx 12284.0$ .
$P \approx 12284$ dollars.

d) (i) Each withdrawal of $5000$ is made at the end of a year. To be able to withdraw it in year $k$ , the money needed today is the amount that, after $k$ years at $2.8 %$ , becomes $5000$ , that is $\frac{5000}{1.02 8 ^{k}}$ .

Summing the present values of the $n$ withdrawals, the minimum initial deposit is $Q = \frac{5000}{1.028} + \frac{5000}{1.02 8 ^{2}} + \dots + \frac{5000}{1.02 8 ^{n}} .$

(ii) Indefinite withdrawals: this is a geometric series with $u_{1} = \frac{5000}{1.028}$ and ratio $r = \frac{1}{1.028}$ , with $∣ r ∣ < 1$ .

$Q = \frac{u _{1}}{1 - r} = \frac{\frac{5000}{1.028}}{1 - \frac{1}{1.028}} = \frac{5000}{1.028 - 1} = \frac{5000}{0.028} \approx 178571.4$ .
To the nearest dollar: $178571$ dollars.

Answer: a) $298468$ dollars; b) proved; c) $P = 12284$ dollars; d) (i) proved; (ii) $Q = 178571$ dollars.

Exercise 85 HardIB · P2 TZ0 November 2016

On the day Mary was born, 1 January 1998, her grandparents invested $x$ dollars in a savings account. From that moment on, they deposited $x$ dollars every month, on the first day of the month. The account offered a fixed interest rate of $0.4 %$ per month. Interest was calculated on the last day of each month and added to the account. Let $A_{n}$ dollars be the amount in the account on the last day of the $n$ th month, just after the interest has been added.

a) Find an expression for $A_{1}$ and show that $A_{2} = 1.00 4^{2} x + 1.004 x$ .

b) (i) Write down a similar expression for $A_{3}$ and one for $A_{4}$ .

(ii) Hence show that the amount in Mary's account on the eve of her 10th birthday is given by $251 (1.00 4^{120} - 1) x$ .

c) Write, in terms of $x$ , an expression for $A_{n}$ on the eve of Mary's 18th birthday, clearly showing the value of $n$ .

d) Mary's grandparents want the account to contain at least $20000$ dollars on the eve of her 18th birthday. Determine the minimum value of the monthly deposit $x$ dollars required to achieve this goal. Give your answer to the nearest whole number of dollars.

e) Immediately after turning 18, Mary decides to invest $15000$ dollars of this money in an account of the same type, earning $0.4 %$ monthly interest. Every year, she withdraws $1000$ dollars on her birthday to buy a gift. Determine how long it will be before there is no money left in the account.

Show answer

Answer: a) $A_{1} = 1.004 x$ ; b) (i) $A_{3} = 1.00 4^{3} x + 1.00 4^{2} x + 1.004 x$ , $A_{4} = 1.00 4^{4} x + 1.00 4^{3} x + 1.00 4^{2} x + 1.004 x$ ; c) $A_{216} = 251 (1.00 4^{216} - 1) x$ ; d) $x = 59$ dollars; e) $28$ years.

Show working and answer

a) At the end of month 1 there are $x$ dollars, and the $0.4 %$ interest is added.

$A_{1} = 1.004 x$ .
On the first day of month 2 another $x$ dollars is deposited, and the interest is calculated at the end of the month on $A_{1} + x$ : $A_{2} = 1.004 (1.004 x + x) = 1.00 4^{2} x + 1.004 x .$

b) (i) The same reasoning is repeated:

$A_{3} = 1.004 (A_{2} + x) = 1.00 4^{3} x + 1.00 4^{2} x + 1.004 x$ .
$A_{4} = 1.004 (A_{3} + x) = 1.00 4^{4} x + 1.00 4^{3} x + 1.00 4^{2} x + 1.004 x$ .

(ii) The eve of the 10th birthday is the last day of month $120$ (10 years $= 120$ months).

$A_{120} = x (1.004 + 1.00 4^{2} + \dots + 1.00 4^{120})$ , a geometric sequence with first term $1.004$ , ratio $1.004$ and $120$ terms.
$A_{120} = x \cdot \frac{1.004 ( 1.00 4 ^{120} - 1 )}{1.004 - 1} = \frac{1.004}{0.004} (1.00 4^{120} - 1) x = 251 (1.00 4^{120} - 1) x$ .

c) 18 years is $18 \cdot 12 = 216$ months, so $n = 216$ : $A_{216} = 251 (1.00 4^{216} - 1) x .$

d) We require $251 (1.00 4^{216} - 1) x \geq 20000$ .

$1.00 4^{216} \approx 2.3685$ , so $251 \cdot 1.3685 \approx 343.5$ and $343.5 x \geq 20000$ .
$x \geq 58.2$ . With $58$ dollars we do not reach $20000$ , so the minimum whole deposit is $x = 59$ dollars.

e) With monthly interest of $0.4 %$ , the annual factor is $r = 1.00 4^{12} \approx 1.04907$ . Let $B_{n}$ be the balance after the $n$ th withdrawal (annual, on each birthday):

$B_{1} = 15000 r - 1000$ , $B_{2} = 15000 r^{2} - 1000 r - 1000$ , ..., $B_{n} = 15000 r^{n} - 1000 \frac{r ^{n} - 1}{r - 1}$ .
We look for when it runs out: $B_{n} = 0 \Rightarrow r^{n} (\frac{1000}{r - 1} - 15000) = \frac{1000}{r - 1}$ .
With $\frac{1000}{r - 1} \approx 20379$ : $r^{n} = \frac{20379}{5379} \approx 3.788$ and $n = \frac{ln 3.788}{ln 1.04907} \approx 27.8$ .
After $27$ withdrawals there is still money ( $B_{27} > 0$ ), but at withdrawal number $28$ there is no longer enough for the $1000$ dollars. The money runs out during year 28.

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