Geometric Progression – SS1 Mathematics Lesson Note (First Term)

A geometric progression (G.P.) is a sequence in which each term is found by multiplying the previous term by the same number, called the common ratio.

• Example: 5, 10, 20, 40 has a first term of 5 and a common ratio of 2.
• nth term: Uₙ = arⁿ⁻¹.

Behavioural objectives

By the end of the lesson, students should be able to:

1. define a geometric progression and find its common ratio;
2. find any term of a G.P. using Uₙ = arⁿ⁻¹;
3. find the sum of a given number of terms of a geometric series;
4. state when a G.P. has a sum to infinity and find that sum.

What is a geometric progression?

A geometric progression is a sequence in which the terms increase or decrease by a common ratio.

• General form: a, ar, ar², ar³, …
• Example: in 5, 10, 20, 40, the first term is 5 and the common ratio is 2, because 10/5 = 40/20 = 2.

What is the notation used in a G.P.?

• a: the first term.
• r: the common ratio.
• n: the number of terms, or the position of a term.
• Uₙ: the nth term.
• Sₙ: the sum of the first n terms.
• S∞: the sum to infinity.

How do you find the common ratio?

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

• Rule: r = U₂ ÷ U₁ = U₃ ÷ U₂ = U₄ ÷ U₃.
• Check: the answer must be the same for every pair of neighbouring terms.
• Example: for 3, 6, 12, 24, r = 6/3 = 12/6 = 2.

What are the types of G.P.?

There are three types of G.P., based on the common ratio: increasing, decreasing and alternating.

How do you find the nth term of a G.P.?

The nth term of a G.P. is Uₙ = arⁿ⁻¹.

It comes from the pattern of the terms:

• 1st term: a.
• 2nd term: a × r = ar.
• 3rd term: a × r × r = ar².
• 4th term: a × r × r × r = ar³.
• 8th term: a multiplied by r seven times = ar⁷.
• nth term: a multiplied by r (n – 1) times = arⁿ⁻¹.

Example 1

Given the G.P. 5, 10, 20, 40, find its (a) 9th term (b) nth term.

1. Write down a = 5 and r = 10/5 = 2.
2. (a) Uₙ = arⁿ⁻¹, so U₉ = 5(2)⁹⁻¹ = 5(2)⁸.
3. 2⁸ = 256, so U₉ = 5 × 256 = 1,280.
4. (b) Uₙ = arⁿ⁻¹ = 5(2)ⁿ⁻¹.

Example 2

Find the 6th term of the G.P. 2, 6, 18, …

1. a = 2 and r = 6/2 = 3.
2. U₆ = 2(3)⁶⁻¹ = 2(3)⁵.
3. 3⁵ = 243, so U₆ = 2 × 243 = 486.

How do you find the sum of a geometric series?

A geometric series is the sum of the terms of a G.P., written a + ar + ar² + ar³ + … + arⁿ⁻¹.

How is the formula for r < 1 worked out?

1. Write the series: S = a + ar + ar² + … + arⁿ⁻¹ … (eqn 1).
2. Multiply through by r: rS = ar + ar² + ar³ + … + arⁿ … (eqn 2).
3. Subtract eqn 2 from eqn 1: S – rS = a – arⁿ.
4. Factorise both sides: S(1 – r) = a(1 – rⁿ).
5. Divide by (1 – r): S = a(1 – rⁿ) / (1 – r).

How is the formula for r > 1 worked out?

1. Use the same eqn 1 and eqn 2 as above.
2. Subtract eqn 1 from eqn 2: rS – S = arⁿ – a.
3. Factorise both sides: S(r – 1) = a(rⁿ – 1).
4. Divide by (r – 1): S = a(rⁿ – 1) / (r – 1).

Which formula should you use?

Example 1

Find the sum of ½ + ¼ + 1/8 + … as far as the 6th term.

1. a = ½ and r = ¼ ÷ ½ = ½, so r < 1.
2. S₆ = ½(1 – (½)⁶) / (1 – ½).
3. (½)⁶ = 1/64, and 1 – ½ = ½.
4. S₆ = ½(1 – 1/64) ÷ ½ = 1 – 1/64.
5. S₆ = 63/64.

Example 2

Find the sum of 1 + 3 + 9 + 27 + … + 729.

1. a = 1, r = 3 and the last term Uₙ = 729. First find n.
2. Uₙ = arⁿ⁻¹, so 729 = 1 × 3ⁿ⁻¹.
3. Write 729 as a power of 3: 729 = 3⁶.
4. So n – 1 = 6 and n = 7.
5. Since r > 1, S₇ = 1(3⁷ – 1) / (3 – 1).
6. 3⁷ = 2,187, so S₇ = 2,186 / 2 = 1,093.

What is the sum of a G.P. to infinity?

The sum to infinity is the value the sum approaches as the number of terms becomes very large.

• Condition: it exists only when -1 < r < 1.
• No sum: when r is 1 or more, or -1 or less, the terms do not get smaller, so there is no sum to infinity.
• Formula: S∞ = a / (1 – r).
• Reason: when -1 < r < 1, rⁿ tends to 0 as n becomes very large.

How is the formula S∞ = a / (1 – r) obtained?

1. Start with Sₙ = a(1 – rⁿ) / (1 – r).
2. Let n become very large, so rⁿ tends to 0.
3. Replace rⁿ with 0: S∞ = a(1 – 0) / (1 – r).
4. Simplify: S∞ = a / (1 – r).

Example 1

Find the sum of 1 + ½ + ¼ + … (a) to 10 terms (b) to 100 terms. Hence deduce the sum to infinity.

1. a = 1 and r = ½.
2. (a) S₁₀ = 1(1 – (½)¹⁰) / (1 – ½) = 2(1 – (½)¹⁰).
3. (½)¹⁰ = 1/1024, which is about 0.001.
4. S₁₀ is about 2(1 – 0.001) = 2 – 0.002 = 1.998.
5. (b) S₁₀₀ = 2(1 – (½)¹⁰⁰).
6. (½)¹⁰⁰ is extremely small, so S₁₀₀ is almost exactly 2.
7. Deduction: the sum approaches 2, and S∞ = 1 / (1 – ½) = 2.

Example 2

Find the sum of 45 + 30 + 20 + … to infinity.

1. a = 45 and r = 30/45 = 2/3.
2. Since -1 < r < 1, the sum to infinity exists.
3. S∞ = a / (1 – r) = 45 / (1 – ⅔).
4. 1 – ⅔ = ⅓, so S∞ = 45 ÷ ⅓.
5. S∞ = 45 × 3 = 135.

What is the difference between a G.P. and an A.P.?

A geometric progression (G.P.) is a sequence in which each term is the previous term multiplied by a common ratio.

An arithmetic progression (A.P.) is a sequence in which each term is the previous term plus a common difference.

Evaluation

1. Find the common ratio and the 6th term of the G.P. 3, 6, 12, …
2. Find the sum of the first 5 terms of the series 1 + 3 + 9 + …
3. Find the sum to infinity of the series 8 + 4 + 2 + …
4. Explain why the series 2 + 4 + 8 + … has no sum to infinity.

Assignment

1. The 3rd term of a G.P. is 360 and the 6th term is 1,215. Find the:
   1. common ratio
   2. first term
   3. sum of the first four terms
2. The first term of a G.P. is 48. Find the common ratio between its terms if its sum to infinity is 36.