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Geometric Sequences and Series

Q: How do I find the common ratio?

Divide any term by the previous one: r = u_n+1u_n. In a genuine geometric sequence this ratio is the same between every pair of consecutive terms.

Q: When does a geometric series have a sum to infinity?

Only when |r| < 1. Then the terms shrink toward zero fast enough for the infinite sum to converge to a1-r. If |r| 1 the terms do not shrink and the sum is infinite.

Q: What’s the difference between the term and sum formulas?

The n-th term is ar^n-1 (one fewer power because the first term is unmultiplied). The sum of n terms uses r^n. Keeping the powers straight is the main thing to watch.

Q: Can the common ratio be negative?

Yes. A negative ratio makes the terms alternate in sign, like 5, -10, 20, with r = -2. The sum to infinity still needs |r| < 1, so e.g. r = - 12 converges.

A geometric sequence multiplies by the same number each time. The surprise is that when that number is small enough, you can add infinitely many terms and still get a finite total — the sum to infinity.

30 min Video by TLMaths Sequences and Series

The big picture

Where arithmetic sequences add a constant, geometric sequences $multiply$ by a constant — the common ratio $r$ . That one change unlocks the mathematics of growth and decay: compound interest, radioactive decay, populations, depreciation. The genuinely mind-bending result is the sum to infinity: add $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots$ forever and you get exactly 1. Understanding $why$ that converges (and when it does not) is the conceptual heart of the topic, and it foreshadows the idea of a limit that calculus is built on.

A-Level Maths D5-02: Geometric Sequences — Finding the nth termWatch the walkthrough, then work through the notes below.

Open on YouTube

What you'll be able to do

Identify a geometric sequence and its common ratio $r$
Use the $n$ -th term formula $a r^{n - 1}$
Use the sum formula for a geometric series
Know when the sum to infinity exists ( $∣ r ∣ < 1$ ) and evaluate it

Multiplying by a constant ratio

A $geometric sequence$ multiplies each term by a fixed $common ratio$ $r$ to get the next. The first term is $a$ . So $2, 6, 18, 54, \dots$ has $a = 2$ and $r = 3$ ; and $16, 8, 4, 2, \dots$ has $r = \frac{1}{2}$ .

To reach the $n$ -th term you multiply by $r$ a total of $(n - 1)$ times — the same “gaps not terms” idea as arithmetic, but with multiplication.

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

Multiply the first term by

r

exactly

(n - 1)

times.

a = 3

r = 2

u_{6} = 3 \times 2^{6 - 1} = 3 \times 32 = 96

Answer

u_{6} = 96

Tip — Find $r$ by dividing any term by the one before it: $r = \frac{u _{n + 1}}{u _{n}}$ . It must be the same everywhere, or the sequence is not geometric.

Summing a geometric series

Adding the terms of a geometric sequence gives a geometric series. There is a neat formula, derived by writing the sum $S_{n}$ , multiplying it by $r$ , and subtracting so that almost everything cancels.

Both forms below are the same; use whichever keeps the arithmetic positive ( $r > 1$ or $r < 1$ ).

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

The

r^{n}

is the whole difficulty — do not confuse it with

r^{n - 1}

a = 3

r = 2

n = 8

2Use

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

S_{8} = \frac{3 ( 2 ^{8} - 1 )}{2 - 1}

= 3 (256 - 1) = 3 \times 255 = 765

Answer

S_{8} = 765

The sum to infinity

Here is the remarkable part. If $∣ r ∣ < 1$ , each term is smaller than the last, and the terms shrink towards zero fast enough that the $infinite$ sum settles on a finite value. As $n \to \infty$ , $r^{n} \to 0$ , so the sum formula collapses to a clean result.

If $∣ r ∣ \geq 1$ the terms do not shrink, the sum grows without bound, and there is no sum to infinity. So the condition $∣ r ∣ < 1$ is not a technicality — it is the difference between a finite answer and infinity.

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

Because

r^{n} \to 0

when

∣ r ∣ < 1

a = 8

r = \frac{1}{2}

, and

∣ r ∣ < 1

so the sum exists.

S_{\infty} = \frac{a}{1 - r} = \frac{8}{1 - \frac{1}{2}} = \frac{8}{\frac{1}{2}} = 16

Answer

S_{\infty} = 16

Why this works

The sum to infinity is your first taste of a $limit$ : infinitely many positive terms adding to a finite total because they vanish quickly enough. The same idea — quantities approaching a value they never quite reach — is the foundation of calculus.

Think like an examiner

The sum uses $r^{n}$ , while the $n$ -th term uses $r^{n - 1}$ — mixing them up is the classic geometric-series slip.
Always check $∣ r ∣ < 1$ before quoting a sum to infinity; state the condition explicitly, as it is often worth a mark.
A “convergent” geometric series is exactly one with $∣ r ∣ < 1$ ; “find the values of $r$ for which the sum exists” means solve $∣ r ∣ < 1$ .
For problems giving two terms, divide them to eliminate $a$ and find $r$ first.

Common misconceptions

Using

r^{n - 1}

in the sum formula.

The sum uses

r^{n}

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

Quoting a sum to infinity when

∣ r ∣ \geq 1

It only exists for

∣ r ∣ < 1

; otherwise the series diverges.

Finding

r

by subtracting terms.

Geometric ratio is found by dividing:

r = \frac{u _{n + 1}}{u _{n}}

Forgetting a negative ratio when terms alternate in sign.

A sequence like

4, - 8, 16, \dots

has

r = - 2

; check the signs.

Geometric formulas

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

n

-th term.

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

Sum of

n

terms.

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

Sum to infinity.

Stretch yourself

A geometric series has second term 6 and fifth term 48. Find the first term $a$ , the common ratio $r$ , and the sum of the first 10 terms.

Hint — Dividing the fifth term by the second removes $a$ and gives $r^{3}$ . Then find $a$ and use the sum formula.

Questions students ask

How do I find the common ratio?

Divide any term by the previous one:

r = \frac{u _{n + 1}}{u _{n}}

. In a genuine geometric sequence this ratio is the same between every pair of consecutive terms.

When does a geometric series have a sum to infinity?

Only when

∣ r ∣ < 1

. Then the terms shrink toward zero fast enough for the infinite sum to converge to

\frac{a}{1 - r}

. If

∣ r ∣ \geq 1

the terms do not shrink and the sum is infinite.

What’s the difference between the term and sum formulas?

The

n

-th term is

a r^{n - 1}

(one fewer power because the first term is unmultiplied). The sum of

n

terms uses

r^{n}

. Keeping the powers straight is the main thing to watch.

Can the common ratio be negative?

Yes. A negative ratio makes the terms alternate in sign, like

5, - 10, 20, \dots

with

r = - 2

. The sum to infinity still needs

∣ r ∣ < 1

, so e.g.

r = - \frac{1}{2}

converges.

Key takeaways

A geometric sequence multiplies by a constant ratio $r$ .
$n$ -th term: $a r^{n - 1}$ ; find $r$ by dividing consecutive terms.
Sum of $n$ terms: $\frac{a ( 1 - r ^{n} )}{1 - r}$ — note the $r^{n}$ .
Sum to infinity $\frac{a}{1 - r}$ exists only when $∣ r ∣ < 1$ .
Convergence of the infinite sum is your first encounter with a limit.

How this fits the course

Build on

Sigma Notation
Exponential Models (later)soon

Leads to

Sigma Notation

Test yourself

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