Exponential Sequences

Sequences that multiply by the same number each time: finding the next term, explaining the rule, and jumping on from a given term.

In the exam for Cambridge IGCSE Maths 0580, Edexcel IGCSE Maths 4MA1, Cambridge O Level Maths 4024 · 15 real past-paper questions below

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Multiply Each Time

In some sequences you multiply by the same number each time, instead of adding. The exam asks for the next term, and often to explain how you found it.

Worked example

1
$3, 12, 48, 192$ — The sequence. The terms grow fast, so check for a multiplier, not a gap.
2
$12 \div 3 = 4, 48 \div 12 = 4$ — Divide each term by the one before it. Every answer is 4, so the multiplier is 4.
3
$192 \div 48 = 4$ — Check the last pair too. It is 4 again, so the rule holds all the way.
4
$192 \times 4 = 768$ — Multiply the last term by the multiplier. The next term is 768.

Rule

To explain, write the rule in words: “multiply the previous term by 4”. That sentence earns the explain mark.

Common mistake

Adding the gaps. The gaps 9, 36, 144 are not equal, so this is not a linear sequence.

Key idea

The story goes that a king offered 1 grain of wheat on the first chess square, 2 on the next, then doubling on all 64 squares. The total is $2^{64} - 1$ : over 18 million million million grains. The tale was written down by Ibn Khallikan in 1256.

Jumping From a Given Term

If the question gives a later term, use it. Do not start again from the beginning.

Worked example

1
$5, 10, 20, 40, \dots$ — The sequence. The question also says the 9th term is 1280.
2
$10 \div 5 = 2$ — Find the multiplier first: divide a term by the one before. It is 2.
3
$9 th term = 1280$ — Start from the given term. Counting up from 5 would take eight steps and invite mistakes.
4
$10 th term = 1280 \times 2 = 2560$ — The 10th term is one step after the 9th, so multiply once.
5
$11 th term = 2560 \times 2 = 5120$ — Each further term is one more multiply. Two terms on means multiply twice.

Halving and Other Rules

The multiplier can be a fraction. Then the terms get smaller, but the method is the same.

Worked example

1
$80, 40, 20, 10$ — The sequence. The terms shrink, so the multiplier is less than 1.
2
$40 \div 80 = \frac{1}{2}$ — Divide a term by the one before. 20 ÷ 40 and 10 ÷ 20 also give ½.
3
$\times \frac{1}{2} = \div 2$ — Multiplying by one half is the same as halving, so each term is half the one before.
4
$10 \times \frac{1}{2} = 5$ — Halve the last term. The next term is 5.

Key idea

Check by dividing each term by the one before. The answer should always be the same.

nth Term: First Term × Multiplier

Exams often ask for the nth term: a rule that gives any term from its position n. For a multiply sequence it is:

Formula

T_{n} = a \times r^{n - 1}

$T_{n}$ is the nth term. $a$ is the first term. $r$ is the multiplier (the common ratio): any term ÷ the one before. $n$ is the position: 1 for the first term, 2 for the second, and so on.

Worked example

1
$6, 18, 54, 162$ — The question: find the nth term.
2
$a = 6$ — The first term is 6.
3
$r = 18 \div 6 = 3$ — Divide a term by the one before to get the multiplier. 54 ÷ 18 = 3 as well.
4
$T_{n} = 6 \times 3^{n - 1}$ — Put a and r into the formula. Leave n as a letter.
5
$n = 1 : 6 \times 3^{0} = 6 \times 1 = 6$ — Check with the first term. Any number to the power 0 is 1, so you get 6. Correct.
6
$n = 2 : 6 \times 3^{1} = 18$ — Check one more term. 18 matches, so the rule is right.

Common mistake

Writing $6 \times 3^{n}$ . Put n = 1 in: it gives 18, not 6. The power is $n - 1$ because the first term has not been multiplied yet.

Key idea

Show your working. A correct power of the multiplier, like $3^{n - 1}$ or $3^{n}$ , can still earn a mark even if the rest is not right.

nth Term With Fractions

Two kinds of nth-term question trip people up: a multiplier that is a fraction, and a first term that is a fraction. The formula still works.

1. Halving: 80, 40, 20, 10, …

Worked example

1
$a = 80$ — The first term is 80.
2
$r = 40 \div 80 = \frac{1}{2}$ — The multiplier is a half, because each term halves.
3
$T_{n} = 80 \times (\frac{1}{2})^{n - 1}$ — Put a and r in the formula. The fraction goes in brackets so the whole of ½ is raised to the power.
4
$n = 3 : 80 \times (\frac{1}{2})^{2} = 80 \times \frac{1}{4} = 20$ — Check the 3rd term. It gives 20. Correct.

2. Starting with a fraction: $\frac{1}{3}, 1, 3, 9, 27, \dots$

Worked example

1
$a = \frac{1}{3}, r = 3$ — 1 ÷ ⅓ = 3 and 3 ÷ 1 = 3, so the multiplier is 3.
2
$T_{n} = \frac{1}{3} \times 3^{n - 1}$ — Put a and r in the formula. This answer is fine as it is.
3
$\frac{1}{3} = 3^{- 1}$ — To tidy it, write ⅓ as a power of 3. A negative power means one over.
4
$T_{n} = 3^{- 1} \times 3^{n - 1} = 3^{n - 2}$ — Same base, so add the powers: −1 + (n − 1) = n − 2.
5
$n = 1 : 3^{- 1} = \frac{1}{3}$ — Check the first term. It gives ⅓. Correct.

Common mistake

Writing $3^{n}$ because the terms are powers of 3. Check n = 1: it gives 3, not ⅓. Always test the first term.

Using a Given nth Term

Sometimes the question gives the nth term and asks for one term. Put the position in for n, then do the power first.

1. Find the 3rd term of $T_{n} = 5 \times 0. 4^{n - 1}$

Worked example

1
$n = 3$ — The 3rd term means n = 3.
2
$T_{3} = 5 \times 0. 4^{3 - 1} = 5 \times 0. 4^{2}$ — Replace n with 3 in the power: 3 − 1 = 2.
3
$0. 4^{2} = 0.16$ — Do the power before multiplying: 0.4 × 0.4 = 0.16.
4
$5 \times 0.16 = 0.8$ — Now multiply by 5. The 3rd term is 0.8.

Common mistake

Doing 5 × 0.4 first and then squaring: $2^{2} = 4$ . The power belongs to 0.4 only, so work it out first.

2. A negative multiplier: find the 4th term of $T_{n} = 2 \times (- 3)^{n - 1}$

Worked example

1
$T_{4} = 2 \times (- 3)^{3}$ — n = 4, so the power is 4 − 1 = 3. Keep −3 in brackets.
2
$(- 3)^{3} = (- 3) \times (- 3) \times (- 3) = - 27$ — Two minuses make plus 9, then times −3 gives −27. An odd power keeps the minus.
3
$2 \times (- 27) = - 54$ — Multiply by 2. The 4th term is −54. The terms go 2, −6, 18, −54: the sign flips each time.

Which Method? Read the Question

Look at how each term changes, and at what the question asks.

You see	You do
Equal gaps	Linear Sequences and Common Difference
Same multiplier	multiply by it again
A later term given	multiply from that term
Find the nth term	$a \times r^{n - 1}$ , then check n = 1
nth term given	put n in, do the power first
Gaps grow by the same amount	Quadratic and Cubic Sequences
Explain your answer	write the rule in words

Key idea

Divide two terms next to each other to find the multiplier.

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