IB Applications and Interpretation SL · Arithmetic and geometric sequences · Section 1 of 5

Sequences, progressions and the nth term

3 worked examples · By Álvaro López

Key idea

A sequence is an ordered list of numbers that follows a rule. The two most important in this course are the arithmetic progression (you always add the same amount) and the geometric progression (you always multiply by the same amount). With the nth-term formula you can calculate any term without writing out the previous ones, and this is the basis of interest, depreciation and loans.

Sequences and terms

A sequence is a list of numbers placed in a specific order. Each number is a term, and is named with a letter and a subscript giving its position: $u_{1}$ is the first term, $u_{2}$ the second and $u_{n}$ the one in position $n$ , with $n \in Z^{+}$ . What matters is the rule linking the terms: if you know it, you can continue the list.

A rule can be given in two ways. With a general term formula, such as $u_{n} = 3 n - 1$ , where you simply substitute $n$ : $u_{1} = 2$ , $u_{2} = 5$ , $u_{3} = 8$ . Or with a recurrence relation, which says how to move from one term to the next, such as $u_{n + 1} = u_{n} + 3$ together with the value of $u_{1}$ . In this section you will see the two simplest rules and how to go from the recurrence to the formula.

Arithmetic progression

For example, $20, 17, 14, 11, \dots$ is arithmetic with $u_{1} = 20$ and $d = - 3$ . To find $d$ it is enough to subtract any two consecutive terms, $d = u_{2} - u_{1}$ , and to check whether a list is arithmetic you must confirm that all these differences are the same.

The general term formula comes from counting how many times $d$ has been added. To get from $u_{1}$ to $u_{n}$ there are $n - 1$ jumps of size $d$ :

The intuition is that of a staircase: the second step is one jump above the first, the third is two jumps above, and so step $n$ is $n - 1$ jumps above. With $u_{1} = 20$ and $d = - 3$ , the 10th term is $u_{10} = 20 + 9 \cdot (- 3) = - 7$ .

Geometric progression

For example, $5, 10, 20, 40, \dots$ has $r = 2$ , and $81, - 27, 9, - 3, \dots$ has $r = - \frac{1}{3}$ . The common ratio is found by dividing two consecutive terms, $r = \frac{u _{2}}{u _{1}}$ , and to recognise a geometric progression you must check that all the quotients are the same.

Now each step is a multiplication, so after $n - 1$ steps we have multiplied $u_{1}$ by $r$ exactly $n - 1$ times:

With $u_{1} = 5$ and $r = 2$ , the 7th term is $u_{7} = 5 \cdot 2^{6} = 320$ . Notice the difference in nature: in the arithmetic case $n$ multiplies $d$ , and in the geometric case $n - 1$ is an exponent. That is why the geometric progression grows (or decreases) much faster.

Worked examples

Example 1 Basic

Recognising and calculating a term

a) The sequence $5, 9, 13, 17, \dots$ is an arithmetic progression. Find its common difference and the 25th term.

b) The sequence $2, 6, 18, 54, \dots$ is a geometric progression. Find its common ratio and the 6th term.

c) State whether the sequence $2, 5, 10, 17$ is arithmetic, geometric or neither.

Solution

a) $d = 9 - 5 = 4$ . With $u_{1} = 5$ : $u_{25} = 5 + 24 \cdot 4 = 5 + 96 = 101$ .

b) $r = \frac{6}{2} = 3$ . With $u_{1} = 2$ : $u_{6} = 2 \cdot 3^{5} = 2 \cdot 243 = 486$ .

c) The differences are $5 - 2 = 3$ , $10 - 5 = 5$ and $17 - 10 = 7$ , which are not equal: it is not arithmetic. The quotients are $\frac{5}{2} = 2.5$ , $\frac{10}{5} = 2$ and $\frac{17}{10} = 1.7$ , which are not equal either: it is not geometric. It is neither.

Example 2 Medium

An arithmetic progression from two terms

In an arithmetic progression, the third term is $17$ and the ninth is $- 7$ .

a) Find the first term and the common difference.

b) Write a formula for $u_{n}$ and find the position of the term $- 35$ .

c) Find the first negative term of the progression.

Solution

a) Write $u_{n} = u_{1} + (n - 1) d$ for each piece of data: ${u_{1} + 2 d u_{1} + 8 d = 17 = - 7$ Subtracting the first equation from the second, $6 d = - 24$ , so $d = - 4$ . In the first, $u_{1} - 8 = 17$ , so $u_{1} = 25$ . Check: $u_{9} = 25 + 8 \cdot (- 4) = - 7$ .

b) $u_{n} = 25 + (n - 1) (- 4) = 29 - 4 n$ . For $u_{n} = - 35$ : $29 - 4 n = - 35$ , that is, $- 4 n = - 64$ and $n = 16$ . The term $- 35$ is in position 16.

c) We need $29 - 4 n < 0$ , that is, $n > \frac{29}{4} = 7.25$ . As $n$ is an integer, the first value is $n = 8$ , and $u_{8} = 29 - 32 = - 3$ . Check: $u_{7} = 1 > 0$ .

Example 3 Hard

The bounces of a ball

A ball is dropped onto the floor and bounces several times. The maximum heights of successive bounces form a geometric progression. The first bounce reaches 120 cm and each bounce reaches $80 %$ of the height of the previous one.

a) Calculate the height of the sixth bounce, to three significant figures.

b) Find the first bounce whose height is less than 10 cm.

Another ball, with a different geometric progression, reaches 60 cm on the second bounce and 12.96 cm on the fifth.

c) Find the common ratio and the height of the first bounce of this second ball.

Solution

a) The ratio is $r = 0.8$ and $u_{1} = 120$ . Then $u_{6} = 120 \cdot 0. 8^{5} = 120 \cdot 0.32768 = 39.3216$ . The height is 39.3 cm.

b) We look for the smallest integer $n$ with $120 \cdot 0. 8^{n - 1} < 10$ . With the GDC you can solve the equation $120 \cdot 0. 8^{x - 1} = 10$ , which gives $x \approx 12.14$ , or look at a table of values. As $n$ must be an integer and the height is decreasing, check the neighbouring integers: $n = 12 : 120 \cdot 0. 8^{11} \approx 10.31 > 10 n = 13 : 120 \cdot 0. 8^{12} \approx 8.25 < 10$ The first bounce below 10 cm is the 13th. (Rounding $12.14$ to the nearest integer would give 12, which is incorrect.)

c) Write $u_{2} = u_{1} r = 60$ and $u_{5} = u_{1} r^{4} = 12.96$ . Dividing the second equation by the first, $u_{1}$ disappears and we get $r^{3} = \frac{12.96}{60} = 0.216 \Rightarrow r = 3 0.216 = 0.6$ Then $u_{1} = \frac{60}{0.6} = 100$ . The first bounce reaches 100 cm and the ratio is $0.6$ (each bounce is $60 %$ of the previous one). Check: $100 \cdot 0. 6^{4} = 12.96$ .

In the exam: marks are awarded for writing $u_{n} = u_{1} r^{n - 1}$ with the data substituted, for setting up the inequality or equation for $n$ and justifying that $n$ is an integer (by checking the neighbouring integers), and for giving the answer with units and to three significant figures unless otherwise stated. In part c) divide the two equations instead of trying to solve blindly.

Common mistakes

Summary

A sequence is an ordered list of terms $u_{1}, u_{2}, \dots, u_{n}$ ; it is defined by a general term formula or by recurrence.
Arithmetic progression: the same difference is always added, $u_{n + 1} = u_{n} + d$ , and $u_{n} = u_{1} + (n - 1) d$ . The difference is found by subtracting consecutive terms.
Geometric progression: the terms are always multiplied by the same ratio, $u_{n + 1} = u_{n} r$ , and $u_{n} = u_{1} r^{n - 1}$ . The ratio is found by dividing consecutive terms and can be negative (the signs alternate).
With two known terms, set up two equations: in the arithmetic case subtract them to obtain $d$ , and in the geometric case divide them to obtain $r^{k}$ .
The GDC can solve the system or the equation and produce tables of values; $n$ is always a positive integer, and in inequalities check the neighbouring integers.

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