A geometric sequence multiplies by the same common ratio r each time. Its nth term is uₙ = arⁿ⁻¹. For 3, 9, 27, …, r = 3 and u₆ = 3 × 3⁵ = 729.
The sum of the first n terms is Sₙ = a(1 − rⁿ)/(1 − r). For 8, 4, 2, …, r = 1/2. When −1 < r < 1, the terms shrink and the series has a sum to infinity: S∞ = a/(1 − r) = 8/(1/2) = 16.
Worked example
Find S₅ for 4, 6, 9, …
- a = 4, r = 1.5.
- S₅ = 4(1 − 1.5⁵)/(1 − 1.5) = 4 × (−6.59375)/(−0.5).
Answer:52.75
Key words
- common ratio (r)
- what each term is multiplied by
- sum to infinity
- the limit of the sum when |r| < 1
- convergent
- has a sum to infinity
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Common Core-referenced: HSA-SSE.B.4