Chapter 08 — Sequences and Series — Part 4: Sum of a Finite Geometric Series

CHAPTER 08: SEQUENCES AND SERIES

PART 4 — SUM OF A FINITE GEOMETRIC SERIES

Learning objectives

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

• Explain the meaning of a geometric series.
• Distinguish between a geometric sequence and a geometric series.
• Derive the formula for the sum of a finite geometric series.
• Calculate the sum when the first term, common ratio and number of terms are known.
• Handle geometric series with positive, negative and fractional common ratios.
• Find an unknown number of terms using the sum formula.
• Solve practical problems involving repeated multiplication and geometric sums.

1. Introduction to geometric series

In Part 2, we studied geometric progressions (GPs). A geometric progression is a sequence in which each term is obtained by multiplying the preceding term by a constant number called the common ratio.

Consider the sequence:

2, 6, 18, 54, 162

Each term is obtained by multiplying the preceding term by 3. Therefore, the common ratio is:

r = 3

When we add the terms, we obtain:

2 + 6 + 18 + 54 + 162 = 242

This addition is called a geometric series.

Remember the difference:

• A geometric sequence is an ordered list of terms whose consecutive-term ratio is constant.
• A geometric series is the sum of terms of a geometric sequence.

2. General form of a geometric series

Let the first term of a GP be a and its common ratio be r.

The geometric progression is:

a, ar, ar², ar³, ..., arⁿ⁻¹

The sum of its first n terms is represented by S_n:

S_n = a + ar + ar² + ar³ + ... + arⁿ⁻¹

We want to develop a formula that allows us to calculate this sum without adding every term individually.

3. Derivation of the finite geometric series formula

Start with the sum:

S_n = a + ar + ar² + ... + arⁿ⁻² + arⁿ⁻¹   (1)

Multiply every term by r:

rS_n = ar + ar² + ar³ + ... + arⁿ⁻¹ + arⁿ   (2)

Subtract equation (2) from equation (1):

S_n − rS_n = a − arⁿ

Factorise both sides:

S_n(1 − r) = a(1 − rⁿ)

Divide both sides by (1 − r), provided r is not equal to 1:

S_n = a(1 − rⁿ)/(1 − r)

First form of the finite geometric series formula

S_n = a(1 − rⁿ)/(1 − r),   r ≠ 1

Where:

• a = first term
• r = common ratio
• n = number of terms
• S_n = sum of the first n terms

4. An alternative form of the formula

The formula may also be written by changing the signs of both the numerator and denominator:

S_n = a(rⁿ − 1)/(r − 1),   r ≠ 1

Both forms are equivalent. Choose the form that makes the calculation easiest, but do not change the sign of only the numerator or only the denominator.

Special case: When r = 1

If the common ratio is 1, every term is equal to the first term:

a + a + a + ... + a

There are n terms, each equal to a. Therefore:

S_n = na,   r = 1

Important: Never substitute r = 1 into the fraction formula. Its denominator becomes zero. Use S_n = na instead.

5. Understanding different common ratios

Common ratio	Behaviour	Example
r greater than 1	Terms increase in magnitude	2, 6, 18, 54, ...
0 less than r less than 1	Positive terms decrease in magnitude	16, 8, 4, 2, ...
r less than −1	Signs alternate and magnitude increases	2, −6, 18, −54, ...
−1 less than r less than 0	Signs alternate and magnitude decreases	8, −4, 2, −1, ...
r = 1	All terms are equal	5, 5, 5, 5, ...
r = −1	Terms alternate between two values	4, −4, 4, −4, ...

6. Worked examples

Example 1: Finding the sum when r is greater than 1

Find the sum of the first 6 terms of:

3 + 6 + 12 + 24 + ...

Solution

Identify the given values:

• a = 3
• r = 2
• n = 6

Use:

S_n = a(1 − rⁿ)/(1 − r)

S₆ = 3(1 − 2⁶)/(1 − 2)

= 3(1 − 64)/(−1)

= 3(−63)/(−1)

S₆ = 189

Answer: 189

Example 2: A fractional common ratio

Find the sum of the first 5 terms of:

32 + 16 + 8 + 4 + ...

Solution

• a = 32
• r = 1/2
• n = 5

S₅ = 32[1 − (1/2)⁵]/[1 − 1/2]

= 32(1 − 1/32)/(1/2)

= 32(31/32) × 2

S₅ = 62

Answer: 62

Example 3: A negative common ratio

Find the sum of the first 6 terms of:

4 − 8 + 16 − 32 + ...

Solution

• a = 4
• r = −2
• n = 6

S₆ = 4[1 − (−2)⁶]/[1 − (−2)]

= 4(1 − 64)/3

= 4(−63)/3

S₆ = −84

Answer: −84

Remember that an even power of a negative number is positive.

Example 4: Using the alternative formula

Find the sum of the first 5 terms of:

2 + 10 + 50 + 250 + ...

Solution

a = 2, r = 5, n = 5.

S₅ = 2(5⁵ − 1)/(5 − 1)

= 2(3125 − 1)/4

= 2 × 3124/4

S₅ = 1562

Answer: 1562

Example 5: When the common ratio is 1

Find the sum of the first 15 terms of:

7 + 7 + 7 + 7 + ...

Solution

Here a = 7, r = 1 and n = 15.

Use the special case:

S_n = na

S₁₅ = 15 × 7

S₁₅ = 105

Answer: 105

Example 6: Finding the number of terms

The sum of the first n terms of the GP 2, 4, 8, 16, ... is 62. Find n.

Solution

a = 2, r = 2 and S_n = 62.

62 = 2(1 − 2ⁿ)/(1 − 2)

62 = 2(2ⁿ − 1)

31 = 2ⁿ − 1

32 = 2ⁿ

32 = 2⁵

n = 5

Answer: 5 terms.

Example 7: Finding an unknown first term

The sum of the first 4 terms of a GP with common ratio 3 is 80. Find the first term.

Solution

Given r = 3, n = 4 and S₄ = 80.

80 = a(1 − 3⁴)/(1 − 3)

80 = a(1 − 81)/(−2)

80 = 40a

a = 2

Answer: The first term is 2.

Example 8: Finding an unknown common ratio

A GP has first term 3. The sum of its first 3 terms is 21. Find its common ratio.

Solution

Given a = 3, n = 3 and S₃ = 21.

Write the first three terms:

3 + 3r + 3r² = 21

Divide by 3:

1 + r + r² = 7

r² + r − 6 = 0

(r + 3)(r − 2) = 0

Therefore:

r = 2 or r = −3

Both values satisfy the given information. Hence there are two possible geometric progressions.

Example 9: A practical application involving repeated growth

A savings account receives a deposit of 10,000 shillings in the first month. The deposit made each month is twice the deposit made in the previous month. Find the total deposited over 6 months.

Solution

The deposits form a GP:

10,000 + 20,000 + 40,000 + ...

Given a = 10,000, r = 2 and n = 6.

S₆ = 10,000(1 − 2⁶)/(1 − 2)

= 10,000(−63)/(−1)

= 630,000

Answer: 630,000 shillings.

Example 10: A practical application involving depreciation

A machine is worth 8,000,000 shillings at the beginning of a period. Its value at the end of each year is 80% of its value at the beginning of that year. Find the total of its values at the beginnings of the first four years.

Solution

The values form a GP:

8,000,000 + 6,400,000 + 5,120,000 + 4,096,000

Here a = 8,000,000, r = 0.8 and n = 4.

S₄ = 8,000,000[1 − (0.8)⁴]/(1 − 0.8)

= 8,000,000(1 − 0.4096)/0.2

= 8,000,000 × 0.5904/0.2

S₄ = 23,616,000

Answer: 23,616,000 shillings.

This is the sum of the four stated annual values, not the machine's value at the end of the fourth year.

7. Geometric series and the nth term

Sometimes a question gives the nth-term formula rather than listing the terms.

Recall:

u_n = arⁿ⁻¹

For example, suppose:

u_n = 5 × 2ⁿ⁻¹

Then:

• First term: u₁ = 5
• Second term: u₂ = 10
• Common ratio: r = 2

Example 11

The nth term of a GP is u_n = 6 × 3ⁿ⁻¹. Find the sum of the first 5 terms.

Solution

From the given formula, a = 6 and r = 3.

S₅ = 6(1 − 3⁵)/(1 − 3)

= 6(1 − 243)/(−2)

= 6(−242)/(−2)

S₅ = 726

Answer: 726

8. Comparing arithmetic and geometric series

Feature	Arithmetic series	Geometric series
Underlying sequence	Arithmetic progression	Geometric progression
Relationship between terms	Constant difference	Constant ratio
nth term	a + (n − 1)d	arⁿ⁻¹
Finite sum	n/2 [2a + (n − 1)d]	a(1 − rⁿ)/(1 − r), r ≠ 1

Examination tip: Before choosing a sum formula, check whether consecutive terms have a constant difference or a constant ratio. Do not confuse the two.

9. Common mistakes to avoid

1. Using the arithmetic sum formula for a GP. Check the relationship between consecutive terms first.
2. Confusing the common ratio with the common difference. The ratio is found by division, not subtraction.
3. Forgetting the power n. The finite GP formula contains rⁿ.
4. Using r = 1 in the fraction formula. Use S_n = na instead.
5. Making a sign error. When r is negative, carefully evaluate the power rⁿ.
6. Changing only one sign in the formula. The two equivalent forms must retain matching numerator and denominator signs.
7. Confusing the number of terms with the last power. A series beginning with a = ar⁰ and ending with arⁿ⁻¹ has n terms.

10. Summary of important formulae

1. nth term of a GP

u_n = arⁿ⁻¹

2. Finite sum, first form

S_n = a(1 − rⁿ)/(1 − r),   r ≠ 1

3. Finite sum, alternative form

S_n = a(rⁿ − 1)/(r − 1),   r ≠ 1

4. Special case when r = 1

S_n = na

11. Practice exercises

Attempt the questions before opening the solutions. Show all working and state the values of a, r and n where appropriate.

Exercise 1

Find the sum of the first 7 terms of:

2 + 4 + 8 + 16 + ...

Show solution

a = 2, r = 2, n = 7.

S₇ = 2(1 − 2⁷)/(1 − 2)

= 2(−127)/(−1)

= 254

Exercise 2

Find the sum of the first 6 terms of:

81 + 27 + 9 + 3 + ...

Show solution

a = 81, r = 1/3, n = 6.

S₆ = 81[1 − (1/3)⁶]/(1 − 1/3)

= 81(728/729)/(2/3)

= 91

Exercise 3

Find the sum of the first 5 terms of:

5 − 10 + 20 − 40 + ...

Show solution

a = 5, r = −2, n = 5.

S₅ = 5[1 − (−2)⁵]/[1 − (−2)]

= 5(1 + 32)/3

= 55

Exercise 4

Find the sum of the first 10 terms of a GP whose first term is 4 and common ratio is 1.

Show solution

Since r = 1, use S_n = na.

S₁₀ = 10 × 4

= 40

Exercise 5

The first term of a GP is 3 and the common ratio is 2. Find the sum of the first 8 terms.

Show solution

S₈ = 3(1 − 2⁸)/(1 − 2)

= 3(−255)/(−1)

= 765

Exercise 6

The sum of the first n terms of the GP 3, 6, 12, 24, ... is 381. Find n.

Show solution

a = 3, r = 2.

381 = 3(1 − 2ⁿ)/(1 − 2)

127 = 2ⁿ − 1

128 = 2ⁿ = 2⁷

n = 7

Exercise 7

The sum of the first 4 terms of a GP is 120. If the common ratio is 2, find the first term.

Show solution

120 = a(1 − 2⁴)/(1 − 2)

120 = 15a

a = 8

Exercise 8

A GP has first term 2 and common ratio −3. Find the sum of the first 4 terms.

Show solution

S₄ = 2[1 − (−3)⁴]/[1 − (−3)]

= 2(1 − 81)/4

= −40

Exercise 9

The nth term of a GP is u_n = 4 × 2ⁿ⁻¹. Find the sum of the first 6 terms.

Show solution

a = 4, r = 2, n = 6.

S₆ = 4(1 − 2⁶)/(1 − 2)

= 4(63)

= 252

Exercise 10

A machine is worth 5,000,000 shillings at the beginning of a year. Its value at the beginning of each following year is 90% of its value at the beginning of the previous year. Find the sum of its values at the beginnings of the first 3 years.

Show solution

a = 5,000,000, r = 0.9, n = 3.

S₃ = 5,000,000[1 − (0.9)³]/(1 − 0.9)

= 5,000,000(0.271)/0.1

= 13,550,000 shillings

Exercise 11

The first term of a GP is 6. The sum of its first 3 terms is 42. Find all possible values of the common ratio.

Show solution

6 + 6r + 6r² = 42

r² + r − 6 = 0

(r + 3)(r − 2) = 0

r = −3 or r = 2

Exercise 12

A geometric progression has first term 8 and common ratio 1/2. Find the sum of its first 8 terms.

Show solution

S₈ = 8[1 − (1/2)⁸]/(1 − 1/2)

= 8(255/256)/(1/2)

= 16 × 255/256

= 255/16 = 15.9375

12. Final lesson check

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

1. What makes a series geometric?
2. How is the common ratio calculated?
3. Why do we multiply the sum by r when deriving the formula?
4. Why must r = 1 be treated separately?
5. What happens to the signs of the terms when r is negative?
6. How do you distinguish a finite geometric sum from an infinite geometric sum?

Key lesson: A finite geometric series is the sum of a specified number of terms of a GP. Its sum can be calculated directly using the first term, common ratio and number of terms.

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Chapter 08 — Sequences and Series | Part 4

Continue to Part 5: Sum to Infinity of a Geometric Series.