Sequence and Series II – SS1 Further Mathematics Lesson Note (First Term)

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

Behavioural objectives

1. Find the nth term of a G.P. using T_n = ar^n-1.
2. Find the number of terms of a G.P.
3. Find the geometric mean of two numbers and insert geometric means between two numbers.
4. Find the sum of the first n terms of a G.P.
5. Find the sum to infinity of a G.P. when it exists.

What is a geometric progression?

A geometric progression is a sequence in which the ratio of any term to the term before it is always the same.

• Common ratio (r): the constant multiplier, found by dividing any term by the term before it.
• First term (a): the starting term of the sequence.
• Dividing: dividing by a number is the same as multiplying by its reciprocal, so r can be a fraction.

How do you find the common ratio?

The common ratio is r = T₂/T₁ = T₃/T₂ = T₄/T₃.

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

The nth term of a G.P. is T_n = ar^n-1.

• T_n: the nth term.
• a: the first term.
• r: the common ratio.
• n: the position of the term, or the number of terms.

Example 1: Find the 9th term of the G.P. 2, -10, 50, …

1. Identify the values: a = 2, r = -10/2 = -5, n = 9.
2. Write the formula: T₉ = ar⁸.
3. Substitute: T₉ = 2(-5)⁸ = 2 × 390625.
4. Answer: T₉ = 781250.

Example 2: Find the number of terms of the G.P. 27, 81, 243, … , 3²⁰

1. Identify the values: a = 27 = 3³, r = 3, T_n = 3²⁰.
2. Substitute: 3²⁰ = 3³ × 3^n-1.
3. Add the powers: 3²⁰ = 3ⁿ⁺².
4. Equate the powers: n + 2 = 20.
5. Answer: n = 18.

Example 3: If 7, x, y, 189 are in G.P., find x and y.

1. Identify the values: a = 7, n = 4, T₄ = 189.
2. Substitute: 7r³ = 189.
3. Simplify: r³ = 189/7 = 27 = 3³, so r = 3.
4. Find x: x = T₂ = ar = 7 × 3 = 21.
5. Find y: y = T₃ = ar² = 7 × 3 × 3 = 63.

What is a geometric mean?

The geometric mean of two numbers x and z is the middle term y that makes x, y, z consecutive terms of a G.P.

How is the formula obtained?

1. For consecutive terms x, y, z, the common ratio is r = y/x = z/y.
2. Cross-multiply: y² = xz.
3. Take the square root: y = √(xz).

• Example: the geometric mean of 4 and 9 is √36 = 6, since 4, 6, 9 has r = 3/2.

How do you insert geometric means between two numbers?

1. Count the terms: if k means are inserted, the G.P. has n = k + 2 terms.
2. Let the first number be a and the last number be T_n.
3. Use T_n = ar^n-1 to find r.
4. Find each missing term by multiplying by r.

Example 1: Insert two geometric means between 12 and 324.

1. The G.P. is 12, x, y, 324, so a = 12, n = 4, T₄ = 324.
2. Substitute: 324 = 12r³.
3. Simplify: r³ = 324/12 = 27, so r = 3.
4. Find x: x = ar = 12 × 3 = 36.
5. Find y: y = ar² = 12 × 9 = 108.
6. Answer: the geometric means are 36 and 108.

Example 2: The 2nd term of a G.P. is 9 and the 4th term is 81. Find r and a.

1. Write the two equations: ar³ = 81 … (i) and ar = 9 … (ii).
2. Divide (i) by (ii): r² = 81/9 = 9.
3. Take the square root: r = ±3.
4. Use (ii): a = 9/r, so a = 3 when r = 3, and a = -3 when r = -3.
5. Answer: a = 3 and r = 3, or a = -3 and r = -3.

How do you find the sum of n terms of a G.P.?

The sum of the first n terms of a G.P. is S_n = a(1 – rⁿ)/(1 – r), where r ≠ 1.

How is the formula obtained?

1. Write the series: S_n = a + ar + ar² + … + ar^n-1 … (i).
2. Multiply by r: rS_n = ar + ar² + ar³ + … + arⁿ … (ii).
3. Subtract (ii) from (i): S_n – rS_n = a – arⁿ.
4. Factorise both sides: S_n(1 – r) = a(1 – rⁿ).
5. Divide by (1 – r) to get the formula.

Which form of the formula should you use?

• Note: both forms give the same answer for any r ≠ 1.

Example 1: The 3rd term of a G.P. is 63 and the 5th term is 567. Find the sum of the first six terms.

1. Write the two equations: ar⁴ = 567 … (i) and ar² = 63 … (ii).
2. Divide (i) by (ii): r² = 567/63 = 9, so r = ±3.
3. Use (ii): a = 63/r² = 63/9 = 7.
4. For r = 3: S₆ = 7(3⁶ – 1)/(3 – 1) = 7(728)/2 = 2548.
5. For r = -3: S₆ = 7(1 – (-3)⁶)/(1 + 3) = 7(-728)/4 = -1274.
6. Answer: S₆ = 2548 for the positive ratio (the usual case in textbooks), or -1274 if r = -3.

Example 2: Find the sum of the first 6 terms of the G.P. 18, 6, 2, …

1. Identify the values: a = 18, r = 6/18 = 1/3, n = 6.
2. Choose the formula for r < 1: S₆ = 18(1 – (1/3)⁶)/(1 – 1/3).
3. Simplify: = 18(1 – 1/729)/(2/3) = (18 × 3 × 728)/(2 × 729).
4. Answer: S₆ = 728/27 = 26.96 (to 2 decimal places).

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

The sum to infinity, S_∞, is the value that the sum of n terms of a G.P. approaches as n becomes very large.

• Formula: S_∞ = a/(1 – r).
• Condition: it exists only when -1 < r < 1.
• Reason: when r is between -1 and 1, rⁿ gets closer to 0 as n grows, so the formula for S_n reduces to a/(1 – r).
• No sum: if r is 1, -1 or beyond these values, as in 2, 4, 8, …, the terms do not shrink and there is no sum to infinity.

Example: Find the sum to infinity of 1, 1/4, 1/16, 1/64, …

1. Identify the values: a = 1, r = (1/4)/1 = 1/4.
2. Check the condition: -1 < 1/4 < 1, so the sum exists.
3. Substitute: S_∞ = 1/(1 – 1/4) = 1/(3/4).
4. Answer: S_∞ = 4/3.

What are the common mistakes with G.P. questions?

• Using n instead of n – 1: the power of r in T_n is always n – 1.
• Losing the negative ratio: when r² is given, remember r = ± the square root.
• Subtracting terms: r is found by dividing, not by subtracting.
• Ignoring the condition: check -1 < r < 1 before using S_∞.

Evaluation

1. Find the 7th term of the G.P. 3, 6, 12, …
2. Insert two geometric means between 8 and 27.
3. Find the sum of the first 5 terms of the G.P. 2, 6, 18, …
4. Find the sum to infinity of the G.P. 12, 4, 4/3, …

Assignment

1. Find the 8th term of the G.P. 5, 10, 20, …
2. The 2nd term of a G.P. is 6 and the 5th term is 162. Find the first term and the common ratio.
3. Find the sum of the first 5 terms of the G.P. 81, 27, 9, …
4. Find the sum to infinity of the G.P. 9, 3, 1, …