Geometric Sequences and Series

Including sum to infinity and sigma notation

SPEC 4.5 • SEQUENCES AND SERIES

Geometric Sequences

Definition

A geometric sequence is a sequence where each term is obtained by multiplying the previous term by a constant value. This constant is called the common ratio, denoted by

r

.

Common Ratio

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

nth Term Formula

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

Key

a

= first term,

r

= common ratio,

n

= term number

Examples: 2, 6, 18, 54, ... has

a = 2

,

r = 3

80, 40, 20, 10, ... has

a = 80

,

r = 0.5

Finding a Specific Term

Worked Example01

A geometric sequence has first term 3 and common ratio 2.

Find the 10th term.

Step 1 Identify the values

a = 3

,

r = 2

,

n = 10

Step 2 Substitute into formula

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

u_{10} = 3 \times 2^{10 - 1} = 3 \times 2^{9} = 3 \times 512 = 1536

Finding a and r from Two Terms

Worked Example02

The 3rd term of a geometric sequence is 36 and the 6th term is 972.

Find the first term and common ratio.

Step 1 Set up equations

u_{3} = a r^{2} = 36

... (1)

u_{6} = a r^{5} = 972

... (2)

Step 2 Divide (2) by (1)

\frac{a r ^{5}}{a r ^{2}} = \frac{972}{36}

, so

r^{3} = 27

, giving

r = 3

Step 3 Substitute into (1):

a \times 9 = 36

, so

a = 4

Practice

1 A geometric sequence has first term 5 and common ratio

\frac{1}{2}

.

(a) Find the 8th term

(b) Which term of the sequence equals

\frac{5}{512}

?

Part a

u_{8} = 5 \times (\frac{1}{2})^{7} = 5 \times \frac{1}{128} = \frac{5}{128}

Part b

5 \times (\frac{1}{2})^{n - 1} = \frac{5}{512}

(\frac{1}{2})^{n - 1} = \frac{1}{512} = (\frac{1}{2})^{9}

So

n - 1 = 9

, giving

n = 10

(10th term)

Geometric Series

Definition

A geometric series is the sum of the terms of a geometric sequence. We use

S_{n}

to denote the sum of the first

n

terms.

Sequence

2, 6, 18, 54, 162

The individual terms

Series

2 + 6 + 18 + 54 + 162 = 242

The sum of the terms

Notation

S_{n} = u_{1} + u_{2} + u_{3} + \dots + u_{n} = a + a r + a r^{2} + \dots + a r^{n - 1}

Deriving the Sum Formula

Consider the sum of a geometric series with first term

a

, common ratio

r

, and

n

terms.

The Multiply and Subtract Method

1 Write out

S_{n}

:

S_{n} = a + a r + a r^{2} + \dots + a r^{n - 1}

2 Multiply by

r

:

r S_{n} = a r + a r^{2} + a r^{3} + \dots + a r^{n}

3 Subtract:

S_{n} - r S_{n} = a - a r^{n}

4 Factor:

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

Alternative Form

Multiplying top and bottom by

- 1

gives:

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

Sum Formulae

Version 1 (when |r| < 1)

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

Version 2 (when |r| > 1)

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

When to Use Which?

Version 1: When

∣ r ∣ < 1

(avoids negative numbers)

Version 2: When

∣ r ∣ > 1

(avoids negative denominator)

Both give the same answer - choose for convenience!

Finding the Sum

Worked Example03

Find the sum of the first 8 terms of the geometric series: 3 + 6 + 12 + 24 + ...

a = 3

,

r = 2

,

n = 8

Using

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

(since

∣ r ∣ > 1

):

S_{8} = \frac{3 ( 2 ^{8} - 1 )}{2 - 1} = \frac{3 ( 256 - 1 )}{1} = 3 \times 255 = 765

Working Backwards from Sum

Worked Example04

A geometric series has first term 4 and common ratio 3.

The sum of the first

n

terms is 4372. Find

n

.

Step 1 Set up equation

\frac{4 ( 3 ^{n} - 1 )}{3 - 1} = 4372

Step 2 Simplify:

2 (3^{n} - 1) = 4372

, so

3^{n} - 1 = 2186

Step 3 Solve:

3^{n} = 2187 = 3^{7}

Therefore

n = 7

Convergence & Sum to Infinity

Convergence Condition

A geometric series converges (has a finite sum to infinity) if and only if

∣ r ∣ < 1

.

When

∣ r ∣ < 1

, as

n \to \infty

,

r^{n} \to 0

.

Convergent (|r| < 1)

Example: $r = 0.5$

Terms: 8, 4, 2, 1, 0.5, ...

Terms approach 0

Divergent (|r| ≥ 1)

Example: $r = 2$

Terms: 3, 6, 12, 24, ...

Terms grow forever

Sum to Infinity (|r| < 1 only)

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

Finding Sum to Infinity

Worked Example05

Find the sum to infinity of the geometric series:

24 + 12 + 6 + 3 + ...

Step 1 Identify:

a = 24

,

r = \frac{12}{24} = \frac{1}{2}

Step 2 Check convergence:

∣ r ∣ = \frac{1}{2} < 1

✓

Step 3 Apply formula:

S_{\infty} = \frac{24}{1 - \frac{1}{2}} = \frac{24}{\frac{1}{2}} = 24 \times 2 = 48

Finding r from Sum to Infinity

Worked Example06

A geometric series has first term 10 and sum to infinity 40.

Find the common ratio.

Step 1 Set up equation:

40 = \frac{10}{1 - r}

Step 2 Solve:

40 (1 - r) = 10

, so

1 - r = \frac{1}{4}

Step 3 Therefore:

r = 1 - \frac{1}{4} = \frac{3}{4}

Check:

∣ r ∣ = \frac{3}{4} < 1

✓ (valid for convergence)

Practice

2 A geometric series has first term 6 and common ratio

- \frac{2}{3}

.

(a) Find the sum to infinity

(b) Find the difference between

S_{\infty}

and

S_{5}

Part a Check:

∣ r ∣ = \frac{2}{3} < 1

✓

S_{\infty} = \frac{6}{1 - ( - \frac{2}{3} )} = \frac{6}{\frac{5}{3}} = 6 \times \frac{3}{5} = \frac{18}{5} = 3.6

Part b

S_{5} = \frac{6 ( 1 - ( - \frac{2}{3} ) ^{5} )}{1 - ( - \frac{2}{3} )} = \frac{6 ( 1 + \frac{32}{243} )}{\frac{5}{3}} = \frac{6 \times \frac{275}{243}}{\frac{5}{3}} \approx 3.95

Difference:

S_{\infty} - S_{5} \approx 3.6 - 3.95 = - 0.35

Identifying Geometric Series

Key Skill

Recognise when a sum in sigma notation represents a geometric series.

How to Identify

A sigma notation represents a geometric series if the general term has the form

a \times r^{(expression in k)}

where

a

and

r

are constants.

Sigma NotationGeometric?Reason

k = 1 \sum n 3 \times 2^{k - 1}

Yes ✓Form

a r^{k - 1}

k = 1 \sum n (3 k + 2)

No ✗Linear (arithmetic)

k = 1 \sum n k \times 2^{k}

No ✗Mixed (k varies)

Evaluating Sigma Notation

Worked Example07

Evaluate

k = 1 \sum 8 5 \times 2^{k - 1}

Step 1 Identify this is geometric

When

k = 1

:

5 \times 2^{0} = 5

, when

k = 2

:

5 \times 2^{1} = 10

So

a = 5

,

r = 2

,

n = 8

Step 2 Apply sum formula:

S_{8} = \frac{5 ( 2 ^{8} - 1 )}{2 - 1} = 5 \times 255 = 1275

Real-World Application

Worked Example08

A ball is dropped from a height of 10 m. Each bounce reaches 80% of the previous height.

Find the total distance travelled by the ball.

Step 1 Initial drop: 10 m down

Bounces: 8 m up + 8 m down, 6.4 m up + 6.4 m down, ...

Step 2 Total =

10 + 2 (8 + 6.4 + 5.12 + ...)

The bracket is GP with

a = 8

,

r = 0.8

Step 3

S_{\infty} = \frac{8}{1 - 0.8} = \frac{8}{0.2} = 40

Step 4 Total =

10 + 2 (40) = 90

metres

Practice

3 Evaluate

k = 1 \sum \infty 12 \times (\frac{1}{4})^{k - 1}

Step 1 Identify:

a = 12

,

r = \frac{1}{4}

Step 2 Check convergence:

∣ r ∣ = \frac{1}{4} < 1

✓

Step 3 Apply sum to infinity formula:

S_{\infty} = \frac{12}{1 - \frac{1}{4}} = \frac{12}{\frac{3}{4}} = 12 \times \frac{4}{3} = 16

Summary

Key Formulae

ConceptFormula

Common ratio

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

nth term

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

Finite sum

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

or

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

Sum to infinity

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

(only when

∣ r ∣ < 1

)

Geometric in sigmaGeneral term is

a \times r^{power}

Key Exam Skills

Find common ratio, nth term, and specific terms

Use finite sum formulae appropriately

Check

∣ r ∣ < 1

before using sum to infinity

Identify geometric series from sigma notation

Apply to real-world problems