About the Geometric Sequence Calculator
This geometric sequence calculator works with any sequence where each term is the previous one multiplied by a fixed common ratio r — for example 3, 6, 12, 24, … (r = 2) or 100, 50, 25, … (r = ½). Enter the first term, the ratio and a term number n to get the nth term, the sum of the first n terms, the infinite sum when it exists, and the list of terms.
It is built for algebra and precalculus students, and for anyone modelling repeated percentage change: bouncing balls, compound growth, drug doses that decay, or a salary rising a fixed percentage each year. The chart and table show each term and the running total so you can see whether the series grows without bound or settles down.
An infinite geometric series converges only when the ratio is strictly between −1 and 1; then its sum is a₁ ÷ (1 − r). Otherwise the calculator reports that the series diverges.
With the default inputs, the nth term (aₙ) is 1,536. Change any value above to recalculate instantly.
How to use the geometric sequence calculator
- 1Enter the first term a₁.
- 2Enter the common ratio r — divide any term by the one before it to find it.
- 3Enter n, the term you want or the number of terms to add.
- 4Read the nth term, the partial sum and, if |r| < 1, the infinite sum.
- 5Use the table to see every term and running total.
Formula and method
Each term is the first term multiplied by the ratio n − 1 times, giving the explicit formula aₙ = a₁rⁿ⁻¹. Multiplying the partial sum Sₙ by r and subtracting it from Sₙ cancels every middle term, leaving Sₙ(1 − r) = a₁(1 − rⁿ); when r = 1 every term is equal and Sₙ = n·a₁.
If −1 < r < 1, rⁿ shrinks toward 0 as n grows, so the partial sums approach a₁ ÷ (1 − r). For |r| ≥ 1 the terms do not shrink and the infinite sum does not exist. Very large results are limited to double precision (about 15 significant digits).
- a₁
- First term of the sequence
- r
- Common ratio (each term ÷ the previous term)
- n
- Position of the term, or how many terms are summed
- Sₙ
- Sum of the first n terms
Worked examples
3, 6, 12, … — the 10th term and sum
a₁₀ = 3 × 2⁹ = 3 × 512 = 1,536. The sum of the first 10 terms is 3(1 − 2¹⁰) ÷ (1 − 2) = 3 × 1,023 = 3,069. Since r = 2, the infinite series diverges.
Halving series 100, 50, 25, …
The 8th term is 100 × 0.5⁷ = 0.78125. The first 8 terms add to 100(1 − 0.5⁸) ÷ 0.5 = 199.21875, approaching the infinite sum 100 ÷ (1 − 0.5) = 200.
Salary rising 3% a year for 20 years
Starting at $50,000 and rising 3% a year, the 20th year’s salary is 50,000 × 1.03¹⁹ ≈ $87,675. Total earnings over 20 years are 50,000(1.03²⁰ − 1) ÷ 0.03 ≈ $1,343,519.
Alternating series with r = −1/3
Terms are 9, −3, 1, −1/3, 1/9, … so a₅ = 1/9 ≈ 0.1111. The infinite sum is 9 ÷ (1 + 1/3) = 6.75.
Frequently asked questions
How do you find the common ratio of a geometric sequence?+
Divide any term by the term before it. In 5, 15, 45, 135 the ratio is 15 ÷ 5 = 3. If the ratios between consecutive terms are not all the same, the sequence is not geometric.
What is the formula for the nth term?+
aₙ = a₁ × rⁿ⁻¹, where a₁ is the first term and r is the common ratio. For 2, 6, 18, … the 7th term is 2 × 3⁶ = 1,458.
When does an infinite geometric series converge?+
Only when the absolute value of the ratio is less than 1 (−1 < r < 1). Then the sum is a₁ ÷ (1 − r). For example 1 + ½ + ¼ + … = 1 ÷ (1 − ½) = 2.
What is the difference between a geometric and an arithmetic sequence?+
An arithmetic sequence adds the same amount each step (2, 5, 8, 11), while a geometric sequence multiplies by the same ratio (2, 6, 18, 54). Arithmetic growth is linear; geometric growth is exponential.
How do I find the number of terms n?+
Rearrange aₙ = a₁rⁿ⁻¹ to n = 1 + log(aₙ ÷ a₁) ÷ log(r). For 3, 6, 12, …, 768 that gives n = 1 + log(256) ÷ log(2) = 9.