Geometric sequences and series practice questions.

A geometric sequence has first term u₁ = 5 and common ratio r = 2. Calculate the sixth term u₆.

• 320
• 80
• 160
• 100

Answer: 160

uₙ = u₁ × rⁿ⁻¹, so u₆ = 5 × 2⁵ = 5 × 32 = 160. Using r⁶ instead of r⁵ gives the distractor 320.

Determine the sum to infinity of the geometric series 12 + 4 + 4/3 + …

• 18
• 24
• 36
• 16

Answer: 18

Here r = 1/3, so S∞ = u₁/(1 − r) = 12/(2/3) = 18. Dividing by r instead of (1 − r) gives 36.

In a geometric sequence, u₂ = 6 and u₅ = 48. Determine the common ratio r.

• 8
• 4
• 3
• 2

Answer: 2

u₅/u₂ = r³ = 48/6 = 8, so r = 2. Forgetting the cube root gives the distractor 8.

State the condition on the common ratio r for an infinite geometric series to converge.

• r < 1
• |r| < 1
• 0 < r < 1
• r ≤ 1

Answer: |r| < 1

The sum to infinity exists only when |r| < 1, i.e. −1 < r < 1. The condition r < 1 alone is not enough, since e.g. r = −2 gives a divergent series.

Calculate the sum of the first five terms of the geometric series 2 + 6 + 18 + …

• 242
• 162
• 484
• 121

Answer: 242

S₅ = u₁(r⁵ − 1)/(r − 1) = 2(3⁵ − 1)/2 = 3⁵ − 1 = 242. The distractor 162 is only the fifth term, not the sum.

A geometric sequence has u₁ = 81 and r = 1/3. Determine which term of the sequence is equal to 1/3.

• The 5th term
• The 7th term
• The 6th term
• The 4th term

Answer: The 6th term

81 × (1/3)ⁿ⁻¹ = 1/3 requires (1/3)ⁿ⁻¹ = 1/243 = (1/3)⁵, so n − 1 = 5 and n = 6.