Geometric Sequences
Multiplying by the same amount each time — an exponential on the integers.
The explanation
A geometric sequence multiplies by a fixed amount, the common ratio r, each step.
3, 6, 12, 24, … has r = 2.
81, 27, 9, 3, … has r = 1/3.
Any term is given by:
aₙ = a₁·r^(n−1)
The 8th term of 3, 6, 12, …: a₈ = 3·2⁷ = 384.
Find r by dividing any term by the one before it. Division, not subtraction — that is the whole distinction from arithmetic.
A negative ratio makes the signs alternate: 2, −6, 18, −54 has r = −3.
A geometric sequence satisfies aₙ = r·aₙ₋₁ with explicit formula aₙ = a₁r^{n−1}. It is an exponential function on a discrete domain, just as an arithmetic sequence is a linear one, and the same n − 1 offset appears for the same reason.
The finite sum is Sₙ = a₁(1 − rⁿ)/(1 − r) for r ≠ 1, derived by subtracting rSₙ from Sₙ so that all interior terms cancel — the same telescoping trick used to convert repeating decimals to fractions.
When |r| < 1, rⁿ → 0 and the infinite series converges to S = a₁/(1 − r). This is why 0.999… = 1: it is the geometric series with a₁ = 0.9 and r = 0.1. For |r| ≥ 1 the series diverges.
Distinguishing sequence types is a matter of testing differences and ratios: constant differences mean arithmetic, constant ratios mean geometric, and neither means some other pattern entirely.
Worked example
For 5, 20, 80, …, find the 7th term. Then find the sum of the infinite series 8 + 4 + 2 + 1 + …
- r = 20/5 = 4, a₁ = 5.
- a₇ = 5·4⁶ = 5(4096) = 20,480.
- Second series: a₁ = 8, r = 1/2, and |r| < 1 so it converges.
- S = 8/(1 − 1/2) = 8/(1/2).
Answer: a₇ = 20,480; the infinite sum is 16.
Common mistakes
- Finding r by subtracting consecutive terms.
- Applying the infinite sum formula when |r| ≥ 1, where the series has no finite sum.