Sequences & Progressions

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Geometric Progressions

⏱ 4 min read🧩 6 question types🎯 13 practice Q

The idea in one minute

A geometric progression (GP) grows or shrinks by a fixed multiplier, the common ratio $r$ . Its $n$ -th term is $ar^{n-1}$ , and a GP with $|r| < 1$ has an infinite sum of $\dfrac{a}{1-r}$ . Doubling money, bounce heights and repeated percentage growth are GPs in disguise.

What a geometric progression is

A geometric progression (GP) is a list where every term comes from the previous one by multiplying the same fixed number.

That number is the common ratio, written $r$ . The first term is $a$ .

3, 6, 12, 24 is a GP: each term is twice the one before. Here $a = 3$ and $r = 2$ .

Doubling money, bounce heights and repeated percentage growth all move by a fixed ratio.

Rule: Check $r$ first: divide any term by the one before it. The answer must be the same everywhere.

A ratio can be a fraction. 8, 4, 2, 1 has $r = \dfrac{1}{2}$ — the list shrinks toward zero.

The nth term

The $n$ -th term is the first term multiplied by $n - 1$ jumps of the ratio.

a_n = a r^{\,n-1}

The 5th term of 3, 6, 12, ... is $3 \times 2^4 = 48$ .

Tip: The power is one less than the term number. Term 5 uses $2^4$ , because term 1 uses $2^0$ .

The sum of n terms

Multiply the first term by ( $r^n - 1$ ), then divide by ( $r - 1$ ):

S_n = \frac{a(r^n - 1)}{r - 1}, \qquad r \ne 1

Sum of 6 terms of 4, 12, 36, ...: $\dfrac{4(3^6 - 1)}{2} = 2 \times 728 = 1456$ .

If $r = 1$ , every term equals $a$ , and the sum is simply $n \times a$ .

Watch: $\dfrac{a(r^n-1)}{r-1}$ and $\dfrac{a(1-r^n)}{1-r}$ are the same sum. Keep the signs consistent.

Sum to infinity

When $r$ lies between $-1$ and 1, the terms shrink and the total settles at one number.

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

$8 + 4 + 2 + 1 + \cdots$ : $a = 8$ , $r = \dfrac{1}{2}$ , so the sum is $\dfrac{8}{1/2} = 16$ .

Watch: This formula fails for $r = 2$ or any ratio of size 1 or more. The list must shrink.

Three-term tricks

Three consecutive GP terms are best named $\dfrac{a}{r}$ , $a$ , $ar$ . Their product is $a^3$ .

Three numbers in GP have sum 21 and product 216. Then $a^3 = 216$ , so $a = 6$ . Now $\dfrac{6}{r} + 6 + 6r = 21$ gives $r = 2$ : the numbers are 3, 6, 12.

The middle of any three terms is the geometric mean of its neighbours: $b^2 = ac$ .

Bounces and doubling

A ball dropped from height $h$ , rebounding to a fraction $r$ each time, travels in total:

h + \frac{2hr}{1 - r}

The first drop happens once; every later up-and-down pair happens twice.

From 36 m, rebounding to half each time: the rebounds are $18 + 9 + 4.5 + \cdots = 36$ . Total $= 36 + 2 \times 36 = 108$ m.

Money that multiplies is a GP too. Savings of Rs 2, 4, 8, ... for 10 months total $2(2^{10} - 1) = 2046$ .

Rs 1 doubled daily for 15 days gives $1 + 2 + 4 + \cdots + 2^{14} = 2^{15} - 1 = 32767$ .

Tip: "Doubles every year or hour" means $r = 2$ . "Grows 10% every year" means $r = 1.1$ — a GP, not an AP.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

nth term of a GP

How to spot it:

A list with a constant multiplier is given and a later term (5th, 10th) is asked.

a_n = a r^{n-1}

Method

Read $a$ and $r$ .
Raise $r$ to the power $n - 1$ .
Multiply by $a$ .

Why it works:

Each step multiplies by $r$ , and reaching term $n$ takes $n - 1$ multiplications.

Try this

Find the 5th term of the GP 3, 6, 12, 24, ...

Show solution

$a = 3$ , $r = 2$ .
$3 \times 2^4 = 48$ .

Answer

Type 2very common2 practice Q

Sum of n terms of a GP

How to spot it:

'Sum of the first 6 terms' of a multiplying list, or 'how many terms give total S'.

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

Method

Read $a$ , $r$ , $n$ .
Compute $r^n$ .
Apply $\dfrac{a(r^n - 1)}{r - 1}$ .

Why it works:

Subtracting $rS_n$ from $S_n$ collapses the list, which gives the closed form.

Try this

Find the sum of the first 6 terms of 4, 12, 36, 108, ...

Show solution

$a = 4$ , $r = 3$ , $n = 6$ .
$\dfrac{4(729 - 1)}{2} = 2 \times 728$ .

Answer

1456

Type 3common3 practice Q

Sum to infinity

How to spot it:

An endless list with $|r| < 1$ (8 + 4 + 2 + ...) and its total is asked — or the total and one of $a$ , $r$ are given.

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

Method

Confirm $|r| < 1$ .
Apply $\dfrac{a}{1-r}$ .
Reverse use: find $a$ or $r$ from the given sum.

Why it works:

As $n$ grows, $r^n$ vanishes for $|r| < 1$ , so the finite sum settles at $\dfrac{a}{1-r}$ .

Try this

Find the sum of the infinite GP 9 + 3 + 1 + 1/3 + ...

Show solution

$a = 9$ , $r = \dfrac{1}{3}$ .
$\dfrac{9}{2/3} = 13.5$ .

Answer

27/2

Type 4common2 practice Q

Three numbers in GP (sum and product given)

How to spot it:

'Three numbers in GP' with their sum and product — the numbers are asked.

Method

Write the terms $\dfrac{a}{r}$ , $a$ , $ar$ .
Product $= a^3$ gives $a$ .
Sum gives $r$ ; keep both roots.

Why it works:

The symmetric naming makes the product collapse to $a^3$ and the sum linear in $r$ .

Try this

Three numbers in GP have sum 21 and product 216. Find the numbers.

Show solution

$a^3 = 216$ , so $a = 6$ .
$\dfrac{6}{r} + 6r = 15$ gives $r = 2$ or $\dfrac{1}{2}$ .

Answer

3, 6, 12

Type 5occasional2 practice Q

Bouncing ball and repeating distances

How to spot it:

A ball or pendulum covers a fixed fraction of its previous path each time; the total distance is asked.

h + \frac{2hr}{1 - r}

Method

First drop: $h$ , once.
Each later up-down pair: twice the rebound chain.
Rebound chain is a GP; sum it for $|r| < 1$ .

Why it works:

After the first fall, every stretch is travelled twice, and the stretch sizes form a GP.

Try this

A ball is dropped from 36 m and rebounds to half its height each time. Find the total distance.

Show solution

Rebounds sum: $\dfrac{18}{1 - 1/2} = 36$ .
Total $= 36 + 2 \times 36 = 108$ m.

Answer

108 m

Type 6common2 practice Q

Doubling money and repeated growth

How to spot it:

Savings double each month, a count doubles each hour, or a value grows by a fixed per cent each period.

Method

Name $a$ and the multiplier $r$ .
One value: $ar^{n-1}$ .
A running total: $\dfrac{a(r^n - 1)}{r - 1}$ .

Why it works:

Any fixed multiplier per period makes the list a GP, so both formulas apply.

Try this

A man saves Rs 2 in the first month, Rs 4 in the second, Rs 8 in the third, and so on. Find his total savings in 10 months.

Show solution

$a = 2$ , $r = 2$ , $n = 10$ .
$2(2^{10} - 1) = 2 \times 1023$ .

Answer

Rs 2046

Formula sheet

nth term

a_n = a r^{n-1}

$r$ is the common ratio; power is $n-1$.

Sum of n terms

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

For $r \ne 1$; if $r = 1$, the sum is $na$.

Sum to infinity

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

Only when $|r| < 1$.

Three GP terms

\frac{a}{r},\ a,\ ar

Their product is $a^3$.

Bouncing ball total

h + \frac{2hr}{1 - r}

$h$ is the drop height, $r$ the rebound fraction.

Shortcuts that save time

⚡ Name three GP terms a/r, a, ar

The product of the three is $a^3$ , so a given product fixes the middle term at once. The sum then fixes $r$ .

Example

Three numbers in GP have sum 21 and product 216. Find them.

Show solution

$a^3 = 216$ , so $a = 6$ .
$\dfrac{6}{r} + 6 + 6r = 21$ gives $r = 2$ .
The numbers are 3, 6, 12.

Answer

3, 6, 12

⚡ Sum to infinity reflex

See an endless list whose ratio sits between $-1$ and 1 — divide the first term by ( $1 - r$ ). One division, no powers.

Example

Find the sum 8 + 4 + 2 + 1 + ...

Show solution

$a = 8$ , $r = \dfrac{1}{2}$ .
$\dfrac{8}{1 - 1/2} = 16$ .

Answer

⚡ Bouncing ball: first drop once, pairs twice

Add the first drop, then twice the sum of all rebounds. The rebound chain is a GP with first term $hr$ .

Example

A ball is dropped from 36 m and rebounds to half its height each time. Find the total distance travelled.

Show solution

Rebounds: $18 + 9 + 4.5 + \cdots = 36$ .
Total $= 36 + 2 \times 36$ .
$= 108$ m.

Answer

108 m

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $a r^n$ for the nth term — at $n = 1$ it must give $a$ , so the term is $a r^{n-1}$ .

Mistake 02

Applying $\dfrac{a}{1-r}$ to every GP — it works only when $|r| < 1$ ; for $r = 2$ the sum simply grows.

Mistake 03

Mixing signs: $\dfrac{a(r^n - 1)}{r - 1}$ and $\dfrac{a(1 - r^n)}{1 - r}$ are the same sum; flipping one sign makes it negative.

Mistake 04

Counting the first drop twice in a bouncing-ball question — the first fall happens once; later up-and-down pairs happen twice.

Mistake 05

Treating 'doubles every year' as an AP — doubling is a GP with $r = 2$ .

Mistake 06

Forgetting the $r = 1$ case — a GP with $r = 1$ is a constant list, and its sum is $n \times a$ .

Quick revision

Read this the night before the exam.

nth term: $ar^{n-1}$ .
Sum of $n$ terms: $\dfrac{a(r^n-1)}{r-1}$ ; if $r = 1$ , it is $na$ .
Sum to infinity: $\dfrac{a}{1-r}$ , only for $|r| < 1$ .
Three GP terms $\dfrac{a}{r}, a, ar$ ; product $= a^3$ .
Ball: total $= h + \dfrac{2hr}{1-r}$ .
'Doubles' means $r = 2$ ; '+10% per period' means $r = 1.1$ .

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

Suggested time 10 min · wrong answers go to your mistake notebook automatically.

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