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Geometric Sequences

Q: How do I tell whether a sequence is arithmetic or geometric?

Check differences and ratios of consecutive terms. A constant difference means arithmetic; a constant ratio means geometric.

Q: Why does the sum to infinity need |r| < 1?

Only then does r^n shrink to zero. Otherwise the terms never become negligible and the sum grows without limit or oscillates.

Q: Can the ratio be negative?

Yes. The terms alternate in sign, and the sum to infinity still exists if -1 < r < 0.

Q: Why do logarithms appear?

When the number of terms is unknown it sits in an exponent, and logarithms are the tool for solving equations with unknown powers.

A ball that rebounds to 70% of its previous height, an investment growing by 4% a year, a population of bacteria doubling every hour: each follows a $geometric sequence$ , where every term is the previous one multiplied by a fixed ratio. Unlike arithmetic sequences, these can grow explosively or shrink towards zero — and when they shrink, their infinite sum can be finite.

40 min Sequences and Series

The big picture

Geometric sequences are exponential functions sampled at whole-number steps, which is why they model compound interest, depreciation and decay so naturally. The key formulas mirror the arithmetic ones — an nth term and a sum — but the sum is derived by a different trick: multiply by $r$ and subtract, so that almost every term cancels. The genuinely new idea is convergence. If $∣ r ∣ < 1$ , the terms shrink fast enough that adding infinitely many of them approaches a limit, $\frac{a}{1 - r}$ . That is the first time in the course you meet an infinite process with a finite answer, and it underpins the binomial expansion for non-integer powers later on. Many exam questions also require logarithms, because the unknown is an exponent.

What you'll be able to do

Find the $n$ th term of a geometric sequence, $u_{n} = a r^{n - 1}$
Prove and use $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$
Determine when a geometric series converges and use $S_{\infty} = \frac{a}{1 - r}$
Solve problems where the unknown is the number of terms, using logarithms

The nth term

A $geometric sequence$ has first term $a$ and $common ratio$ $r$ : $u_{n + 1} = r u_{n}$ . Its terms are $a, a r, a r^{2}, a r^{3}, \dots$

Getting to the $n$ th term takes $n - 1$ multiplications by $r$ , so $u_{n} = a r^{n - 1}$ .

If $r > 1$ the terms grow; if $0 < r < 1$ they decay towards zero; if $r < 0$ the signs alternate. The ratio is found by dividing any term by the one before it.

u_{n} = a r^{n - 1}

nth term of a geometric sequence.

a r = 12

and

a r^{4} = 96

2Divide:

r^{3} = 8

, so

r = 2

3Then

a = 12 \div 2

Answer

a = 6

r = 2

Tip — Divide the two equations rather than subtracting — the $a$ cancels and leaves a power of $r$ .

The sum of a geometric series

Write $S_{n} = a + a r + a r^{2} + \dots + a r^{n - 1}$ , then multiply by $r$ : $r S_{n} = a r + a r^{2} + \dots + a r^{n}$ .

Subtracting, everything cancels except the first term of $S_{n}$ and the last term of $r S_{n}$ : $S_{n} - r S_{n} = a - a r^{n}$ .

So $S_{n} (1 - r) = a (1 - r^{n})$ , giving the formula below, valid for any $r \neq = 1$ . The equivalent form $\frac{a ( r ^{n} - 1 )}{r - 1}$ avoids negatives when $r > 1$ .

S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r} = \frac{a ( r ^{n} - 1 )}{r - 1}

Sum of the first n terms.

1First term (at

r = 1

a = 6

; ratio

2

; 10 terms.

S_{10} = \frac{6 ( 2 ^{10} - 1 )}{2 - 1} = 6 \times 1023

Answer

6138

Why this works

OCR can ask you to prove this formula. The whole proof is the "multiply by $r$ and subtract" step — write both lines out so the cancellation is visible.

Sum to infinity

If $∣ r ∣ < 1$ , then $r^{n} \to 0$ as $n \to \infty$ , so $S_{n} \to \frac{a}{1 - r}$ . The series is said to $converge$ , and this limit is its $sum to infinity$ .

If $∣ r ∣ \geq 1$ the terms do not shrink to zero, and the series $diverges$ — it has no sum to infinity.

Recurring decimals are geometric series in disguise: $0. \dot{4} = 0.4 + 0.04 + 0.004 + \dots$ with $a = 0.4$ , $r = 0.1$ , giving $\frac{0.4}{0.9} = \frac{4}{9}$ .

S_{\infty} = \frac{a}{1 - r}, ∣ r ∣ < 1

Sum to infinity.

1Initial drop: 5 m.

2After that, each bounce goes up and down: heights

3, 1.8, 1.08, \dots

, a geometric series with

a = 3

r = 0.6

3Sum of heights:

\frac{3}{1 - 0.6} = 7.5

m, travelled twice (up and down).

4Total:

5 + 2 \times 7.5

Answer

20

Tip — In a sum-to-infinity question, state the condition $∣ r ∣ < 1$ explicitly. If $r$ is unknown, the answer must respect it — reject any value of $r$ outside $- 1 < r < 1$ .

Unknown number of terms

When the unknown is $n$ , you end up with $r^{n}$ equal to some number, and logarithms bring the power down: $n = \frac{lo g k}{lo g r}$ .

If $0 < r < 1$ , then $lo g r < 0$ , so dividing by it $reverses$ an inequality. This is the most common algebraic slip in these questions.

1Value after

n

years:

18000 \times 0.8 5^{n}

2Need

0.8 5^{n} < \frac{5000}{18 000} = 0.2 \dot{7}

3Take logs:

n lo g 0.85 < lo g 0.2 \dot{7}

. Dividing by

lo g 0.85 < 0

reverses the sign:

n > \frac{lo g 0.2 7 ˙}{lo g 0.85} = 7.88

4Check: after 7 years the value is

18000 \times 0.8 5^{7} \approx 5770

pounds; after 8 years it is

18000 \times 0.8 5^{8} \approx 4905

pounds.

AnswerAfter 8 years.

Think like an examiner

Divide equations to eliminate $a$ when two terms are given.
State $∣ r ∣ < 1$ whenever you use a sum to infinity.
Show the log step and the inequality reversal when solving for $n$ .
Be ready to prove $S_{n}$ by multiplying by $r$ and subtracting.

Common misconceptions

Using

u_{n} = a r^{n}

The first term is

a r^{0}

, so

u_{n} = a r^{n - 1}

Using

S_{\infty}

with

r = 1.2

A sum to infinity exists only if

∣ r ∣ < 1

Not reversing an inequality after dividing by

lo g 0.85

lo g r < 0

for

0 < r < 1

, so the inequality flips.

Starting a sigma sum’s first term at

r = 0

when the index starts at 1.

Substitute the starting index to find the first term.

Geometric sequences

u_{n} = a r^{n - 1}

nth term.

S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}

Sum of n terms.

S_{\infty} = \frac{a}{1 - r}

Sum to infinity,

∣ r ∣ < 1

Stretch yourself

A geometric series has sum to infinity $36$ , and its second term is $8$ . Find the two possible values of the common ratio, and the first term in each case.

Hint — Write $\frac{a}{1 - r} = 36$ and $a r = 8$ . Eliminate $a$ to get a quadratic in $r$ , and check each root against $∣ r ∣ < 1$ .

Questions students ask

How do I tell whether a sequence is arithmetic or geometric?

Check differences and ratios of consecutive terms. A constant difference means arithmetic; a constant ratio means geometric.

Why does the sum to infinity need

∣ r ∣ < 1

Only then does

r^{n}

shrink to zero. Otherwise the terms never become negligible and the sum grows without limit or oscillates.

Can the ratio be negative?

Yes. The terms alternate in sign, and the sum to infinity still exists if

- 1 < r < 0

Why do logarithms appear?

When the number of terms is unknown it sits in an exponent, and logarithms are the tool for solving equations with unknown powers.

Key takeaways

Geometric: constant ratio $r$ ; $u_{n} = a r^{n - 1}$ .
$S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$ , proved by multiplying by $r$ and subtracting.
If $∣ r ∣ < 1$ the series converges to $S_{\infty} = \frac{a}{1 - r}$ .
Use logarithms to find an unknown number of terms, reversing inequalities when dividing by a negative log.
Geometric sequences model compound growth and decay.

How this fits the course

Build on

Arithmetic Sequences

Exponential Functions

Leads to

Binomial Expansion

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Real past-paper questions on Geometric Sequences, marked mark-by-mark. How you do feeds straight into your weak-topic list, so your revision keeps targeting what actually needs work.

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Geometric Sequences

The nth term

The sum of a geometric series

Sum to infinity

Unknown number of terms

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Geometric Sequences

The nth term

The sum of a geometric series

Sum to infinity

Unknown number of terms

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