Geometric Sequence Calculator
Choose what to find and enter the other values. The nth term, the finite sum, the infinite sum when it exists, and the terms themselves are shown.
Result
0.001953125
Worked step
aₙ = a₁ · rⁿ⁻¹ = 1 × 0.5^9 = 0.001953125
nth term aₙ
0.001953125
Sum of the first n terms Sₙ
1.998046875
Sum to infinity S∞
2
First term a₁
1
Common ratio r
0.5
n
10
Explicit formula
aₙ = 1 × 0.5ⁿ⁻¹
How it works
A geometric sequence multiplies each term by the same number, the common ratio r, to get the next: 1, ½, ¼, ⅛, … has a₁ = 1 and r = ½, while 3, 6, 12, 24, … has r = 2. The nth term is the first term multiplied by r a total of n − 1 times: aₙ = a₁ · rⁿ⁻¹.
The sum of the first n terms, Sₙ, comes from a telescoping trick: multiply the sum by r, subtract, and almost everything cancels, leaving Sₙ = a₁(1 − rⁿ) ÷ (1 − r). When r = 1 every term is the same and the sum is simply n · a₁.
When the ratio is between −1 and 1, rⁿ shrinks towards zero as n grows, so the partial sums settle on a limit: the sum to infinity S∞ = a₁ ÷ (1 − r). This is why 1 + ½ + ¼ + ⅛ + … adds up to exactly 2 and why 0.999… equals 1. When |r| ≥ 1 the terms do not shrink and the series diverges.
You can also solve for the ratio from the first and nth terms, or find which term a given value is. The table lists the first terms with their running sums so you can watch the series converge or grow.
Formula
aₙ = a₁ · rⁿ⁻¹ Sₙ = a₁(1 − rⁿ) ÷ (1 − r) for r ≠ 1; Sₙ = n · a₁ for r = 1 S∞ = a₁ ÷ (1 − r) for |r| < 1 only r = (aₙ ÷ a₁)^(1 ÷ (n − 1)) n = log(aₙ ÷ a₁) ÷ log(r) + 1
Example
a₁ = 1, r = ½, n = 10: a₁₀ = 1 × (½)⁹ = 1 ÷ 512 = 0.001953125, and S₁₀ = (1 − (½)¹⁰) ÷ (1 − ½) = 2 × 1023 ÷ 1024 = 1.998046875. The sum to infinity is 1 ÷ (1 − ½) = 2.
Doubling from 1 for ten terms (r = 2): a₁₀ = 2⁹ = 512 and S₁₀ = 2¹⁰ − 1 = 1,023. The series diverges, so no infinite sum is reported.
Which term of 2, 6, 18, … is 162? n = log(162 ÷ 2) ÷ log(3) + 1 = log(81) ÷ log(3) + 1 = 5.
Assumptions and limitations
- The first term cannot be zero (every term would be zero and no ratio could be recovered). The ratio may be zero or negative; a negative ratio alternates signs.
- The sum to infinity exists only when |r| < 1; otherwise the output says the series diverges.
- Finding n requires a positive ratio other than 1 and an nth term with the same sign as the first term, and only succeeds when the value entered is actually a term of the sequence.
- Finding r from a₁, aₙ and n takes the real root. When n is odd and aₙ ÷ a₁ is positive, both r and −r satisfy the data; the positive one is reported and the step notes the other. When n is odd and the quotient is negative there is no real ratio.
- The table lists at most the first 25 terms. Terms beyond about 1.8 × 10³⁰⁸ cannot be represented and are reported as too large. Results are cleaned to 15 significant digits and shown to 10 decimal places.
Frequently asked questions
Why does 1 + ½ + ¼ + … equal exactly 2 and not just get close?
The partial sums are 1, 1.5, 1.75, 1.875, …, always 2 minus the last term added, and that gap (½)ⁿ⁻¹ can be made smaller than any positive number by taking enough terms. The limit of the partial sums, which is what an infinite sum means, is therefore exactly 2 = a₁ ÷ (1 − r).
Is compound interest a geometric sequence?
Yes: a balance that grows by a fixed percentage each period is a geometric sequence with r = 1 + rate, and a series of equal deposits adds up to a geometric sum. For money questions, use the finance calculators, which handle the deposit timing and rate conventions.
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