Chapter 08 — Sequences and Series (Part 2): Geometric Progressions

CHAPTER 08: SEQUENCES AND SERIES

PART 2 — GEOMETRIC PROGRESSIONS

A-Level Mathematics | Mathematics HipHop

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Learning Objectives

By the end of this lesson, you should be able to:

• Explain the meaning of a geometric progression.
• Identify the common ratio of a GP.
• Distinguish between arithmetic and geometric progressions.
• Find the nth term of a GP.
• Determine missing terms in a GP.
• Find the first term or common ratio when other terms are known.
• Solve problems involving geometric progressions.
• Recognise the behaviour of a GP for different values of its common ratio.

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1. Introduction to Geometric Progressions

In Part 1, we studied arithmetic progressions, where the difference between consecutive terms is constant.

We now consider another important type of sequence called a geometric progression (GP).

A geometric progression is a sequence in which each term after the first is obtained by multiplying the preceding term by a fixed constant.

This fixed constant is called the common ratio and is usually represented by the letter r.

Consider the sequence:

2, 6, 18, 54, 162, ...

Observe that:

• 6 = 2 × 3
• 18 = 6 × 3
• 54 = 18 × 3
• 162 = 54 × 3

Every term is obtained by multiplying the preceding term by 3.

Therefore, this is a geometric progression with common ratio r = 3.

Key idea

An arithmetic progression has a constant difference.

A geometric progression has a constant ratio.

2. Finding the Common Ratio

The common ratio is found by dividing any term by the term immediately before it.

For a GP with consecutive terms u₁, u₂, u₃, ...:

r = (following term) ÷ (preceding term)

In symbols:

r = u₂/u₁ = u₃/u₂ = u₄/u₃

All these ratios must be equal for a non-zero geometric progression.

Worked Example 1

Find the common ratio of the sequence:

5, 15, 45, 135, 405, ...

Solution

Divide the second term by the first:

r = 15/5 = 3.

Check using the next pair:

r = 45/15 = 3.

The ratio is constant.

Common ratio, r = 3.

Worked Example 2

Find the common ratio of:

64, 32, 16, 8, 4, ...

Solution

r = 32/64 = 1/2.

Checking:

16/32 = 1/2.

Common ratio, r = 1/2.

3. Positive, Negative and Fractional Common Ratios

The common ratio may be positive, negative, greater than one, or a fraction.

(a) Positive ratio greater than one

2, 6, 18, 54, 162, ...

Here r = 3. The magnitudes of the terms increase.

(b) Positive ratio between zero and one

80, 40, 20, 10, 5, ...

Here r = 1/2. The terms decrease in magnitude towards zero.

(c) Negative common ratio

3, −6, 12, −24, 48, ...

Here r = −2. The signs alternate, while the magnitudes increase.

(d) Negative fractional common ratio

16, −8, 4, −2, 1, ...

Here r = −1/2. The signs alternate, while the magnitudes decrease.

(e) Common ratio equal to one

7, 7, 7, 7, 7, ...

Here r = 1. Every term is equal.

Remember

A negative common ratio causes successive non-zero terms to alternate in sign. A fractional ratio with magnitude less than one causes the magnitudes to decrease.

4. Arithmetic Progression Compared with Geometric Progression

It is important not to confuse these two types of sequences.

Feature	AP	GP
Rule	Add a constant	Multiply by a constant
Constant	Difference d	Ratio r
Example	2, 5, 8, 11	2, 6, 18, 54

Worked Example 3

Determine whether each sequence is an AP, a GP, or neither.

(a) 4, 8, 12, 16, ...

(b) 3, 9, 27, 81, ...

(c) 2, 5, 10, 17, ...

Solution

(a) Differences are 4, 4, 4. It is an AP.

(b) Ratios are 3, 3, 3. It is a GP.

(c) Differences are 3, 5, 7, while ratios are not constant. It is neither.

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5. The General Form of a Geometric Progression

Let a be the first term and r be the common ratio.

The terms of a GP are:

a, ar, ar², ar³, ar⁴, ...

Notice how the powers of r increase by one for every new term.

Position	Term
1	a
2	ar
3	ar²
4	ar³
5	ar⁴
n	arⁿ⁻¹

6. Deriving the nth Term of a GP

Consider the GP:

a, ar, ar², ar³, ar⁴, ...

The first term contains no factor of r, which is equivalent to a × r⁰.

The second term contains one factor of r.

The third term contains two factors of r.

The fourth term contains three factors of r.

Therefore, the term in position n contains n − 1 factors of r.

THE nth TERM OF A GP

uₙ = arⁿ⁻¹

Where:

• a = first term.
• r = common ratio.
• n = position of the required term.
• uₙ = term in position n.

This formula allows us to find any term without writing all the terms before it.

7. Finding a Particular Term

Worked Example 4

Find the 8th term of the GP:

3, 6, 12, 24, ...

Solution

First term, a = 3.

Common ratio, r = 6/3 = 2.

Required position, n = 8.

Using uₙ = arⁿ⁻¹:

u₈ = 3 × 2⁽⁸⁻¹⁾

= 3 × 2⁷

= 3 × 128

u₈ = 384.

Worked Example 5

Find the 7th term of:

128, 64, 32, 16, ...

Solution

a = 128 and r = 64/128 = 1/2.

u₇ = 128(1/2)⁶

= 128/64

u₇ = 2.

Worked Example 6

Find the 6th term of:

2, −6, 18, −54, ...

Solution

a = 2.

r = −6/2 = −3.

u₆ = 2(−3)⁵

= 2 × (−243)

u₆ = −486.

8. Finding the nth-Term Formula

Sometimes a question gives the first few terms and asks for a formula for the nth term.

Worked Example 7

Find the nth-term formula for:

5, 10, 20, 40, 80, ...

Solution

First term: a = 5.

Common ratio: r = 10/5 = 2.

Using uₙ = arⁿ⁻¹:

uₙ = 5 × 2⁽ⁿ⁻¹⁾.

Required formula: uₙ = 5 × 2⁽ⁿ⁻¹⁾.

Worked Example 8

Find the nth term of:

81, 27, 9, 3, 1, ...

Solution

a = 81.

r = 27/81 = 1/3.

Therefore:

uₙ = 81(1/3)⁽ⁿ⁻¹⁾.

Required formula: uₙ = 81(1/3)⁽ⁿ⁻¹⁾.

9. Finding Missing Terms

In a GP, the ratio between consecutive terms remains constant. This allows us to calculate missing terms.

Worked Example 9

Find the missing terms:

3, __, __, 81, ...

Solution

Let the common ratio be r.

The first term is 3.

The fourth term is 81.

Using the nth-term formula:

u₄ = ar³

81 = 3r³

r³ = 27

r = 3.

The missing terms are:

3, 9, 27, 81, ...

Worked Example 10

Insert two positive geometric means between 4 and 108.

Solution

Let the four terms be:

4, 4r, 4r², 4r³.

The fourth term is 108:

4r³ = 108

r³ = 27

r = 3.

Therefore, the two inserted terms are:

4 × 3 = 12

4 × 3² = 36

Answer: 12 and 36.

Geometric means

Numbers inserted between two positive numbers to form a GP are called geometric means.

10. Finding the First Term and Common Ratio from Given Terms

We can use simultaneous equations involving powers of the common ratio.

Worked Example 11

The second term of a GP is 12 and the fifth term is 324. Find the first term and common ratio, given that the common ratio is positive.

Solution

Let the first term be a and the common ratio be r.

Second term:

ar = 12   (1)

Fifth term:

ar⁴ = 324   (2)

Divide equation (2) by equation (1):

r³ = 324/12

r³ = 27

r = 3.

Substitute into equation (1):

3a = 12

a = 4.

First term = 4; common ratio = 3.

Worked Example 12

The third term of a GP is 20 and the sixth term is 160. Find the positive common ratio and the first term.

Solution

Third term: ar² = 20   (1)

Sixth term: ar⁵ = 160   (2)

Divide equation (2) by equation (1):

r³ = 160/20 = 8

r = 2.

Substitute into equation (1):

a(2²) = 20

4a = 20

a = 5.

First term = 5; common ratio = 2.

11. Determining Whether a Number Is a Term of a GP

To check whether a given number belongs to a GP, equate the nth-term formula to that number and solve for n.

The resulting value of n must be a positive integer.

Worked Example 13

Determine whether 486 is a term of the GP:

2, 6, 18, 54, ...

Solution

a = 2 and r = 3.

uₙ = 2 × 3⁽ⁿ⁻¹⁾.

Set uₙ = 486:

2 × 3⁽ⁿ⁻¹⁾ = 486

3⁽ⁿ⁻¹⁾ = 243

243 = 3⁵.

Therefore:

n − 1 = 5

n = 6.

Yes. 486 is the 6th term.

12. Applications of Geometric Progressions

Geometric progressions can describe situations where a quantity repeatedly changes by the same multiplying factor.

Examples include:

• A quantity that doubles at each stage.
• A population model with a fixed percentage growth rate.
• The repeated reduction of a quantity by a fixed percentage.
• The successive sizes of an object after repeated scaling.

Worked Example 14 — Repeated Growth

A school has 200 learners. The number increases by 10% each year. Assuming this rate continues, find the estimated number after 4 years.

Solution

Initial number = 200.

Growth factor = 1 + 10/100 = 1.1.

After each year, the number is multiplied by 1.1.

After 4 years:

Number = 200 × (1.1)⁴

= 200 × 1.4641

Estimated number = 292.82, approximately 293 learners.

Worked Example 15 — Repeated Reduction

A machine's value is 8,000,000 shillings. It loses 20% of its value each year. Find its value after 3 years.

Solution

Initial value = 8,000,000.

Remaining fraction each year = 1 − 0.20 = 0.8.

After 3 years:

Value = 8,000,000 × (0.8)³

= 8,000,000 × 0.512

Value = 4,096,000 shillings.

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13. Summary of Important Formulae

General form of a GP:

a, ar, ar², ar³, ...

Common ratio:

r = following term / preceding term

nth term:

uₙ = ar⁽ⁿ⁻¹⁾

Repeated percentage increase:

New amount = initial amount × (1 + p/100)ⁿ

Repeated percentage decrease:

New amount = initial amount × (1 − p/100)ⁿ

14. Practice Exercises

Attempt all questions before opening the answers. Show every important step in your working.

Question 1

Define a geometric progression and explain what is meant by its common ratio.

Question 2

Find the common ratio of each sequence:

(a) 4, 12, 36, 108, ...

(b) 64, 32, 16, 8, ...

(c) 5, −10, 20, −40, ...

(d) 3, 3, 3, 3, ...

Question 3

Determine whether each sequence is an AP, a GP, or neither:

(a) 6, 10, 14, 18, ...

(b) 2, 8, 32, 128, ...

(c) 1, 4, 9, 16, ...

Question 4

Find the nth-term formula for:

(a) 7, 21, 63, 189, ...

(b) 96, 48, 24, 12, ...

(c) 2, −4, 8, −16, ...

Question 5

Find the 9th term of the GP:

2, 6, 18, 54, ...

Question 6

Find the 8th term of a GP whose first term is 256 and common ratio is 1/2.

Question 7

Find the missing terms:

5, __, __, 135, ...

Assume the common ratio is positive.

Question 8

Insert two positive geometric means between 3 and 81.

Question 9

The second term of a GP is 18 and the fifth term is 486. Find its first term and positive common ratio.

Question 10

Determine whether 1,024 is a term of the GP:

4, 8, 16, 32, ...

Question 11

A quantity begins at 500 and increases by 5% every year. Find its value after 3 years.

Question 12

A vehicle is valued at 12,000,000 shillings. Its value decreases by 10% each year. Find its value after 2 years.

Open Answers and Worked Solutions

Question 1

A GP is a sequence in which each term after the first is obtained by multiplying the preceding term by a constant number called the common ratio.

Question 2

(a) r = 12/4 = 3.

(b) r = 32/64 = 1/2.

(c) r = −10/5 = −2.

(d) r = 3/3 = 1.

Question 3

(a) AP, because the difference is always 4.

(b) GP, because the ratio is always 4.

(c) Neither, because neither the differences nor ratios are constant.

Question 4

(a) a = 7, r = 3.

uₙ = 7 × 3⁽ⁿ⁻¹⁾.

(b) a = 96, r = 1/2.

uₙ = 96(1/2)⁽ⁿ⁻¹⁾.

(c) a = 2, r = −2.

uₙ = 2(−2)⁽ⁿ⁻¹⁾.

Question 5

a = 2, r = 3.

u₉ = 2 × 3⁸ = 2 × 6561.

Answer: 13,122.

Question 6

u₈ = 256(1/2)⁷ = 256/128.

Answer: 2.

Question 7

u₄ = ar³.

135 = 5r³.

r³ = 27, so r = 3.

Answer: 5, 15, 45, 135, ...

Question 8

Let the terms be 3, 3r, 3r², 3r³.

3r³ = 81, so r³ = 27 and r = 3.

Answer: 9 and 27.

Question 9

ar = 18 and ar⁴ = 486.

Divide: r³ = 486/18 = 27.

r = 3.

a = 18/3 = 6.

Answer: a = 6, r = 3.

Question 10

uₙ = 4 × 2⁽ⁿ⁻¹⁾.

Set equal to 1,024:

4 × 2⁽ⁿ⁻¹⁾ = 1,024 = 2¹⁰.

Since 4 = 2², we get 2 + n − 1 = 10.

n = 9.

Answer: Yes, 1,024 is the 9th term.

Question 11

Growth factor = 1.05.

Value after 3 years = 500 × 1.05³.

= 500 × 1.157625.

Answer: 578.8125.

Question 12

Remaining factor = 0.9.

Value after 2 years = 12,000,000 × 0.9².

= 12,000,000 × 0.81.

Answer: 9,720,000 shillings.

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15. Final Learning Check

Before moving to the next part, make sure you can:

• Recognise a geometric progression.
• Calculate its common ratio correctly.
• Derive and apply the nth-term formula.
• Find missing terms and geometric means.
• Use two given terms to calculate a and r.
• Solve repeated percentage growth and reduction problems.

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