Sum of Series Calculator

Calculate the sum of an arithmetic or geometric series (Sₙ), the nth term, and the average term from the first term, common difference or ratio, and number of terms.

Quick Facts

Arithmetic sum
Sₙ = n/2 × (2a₁ + (n−1)d)
Equivalent to n × the average of the first and last term.
Geometric sum
Sₙ = a₁(1−rⁿ)/(1−r), r ≠ 1
When r = 1, every term equals a₁, so Sₙ = a₁ × n.
Infinite geometric sum
S = a₁/(1−r), only if |r| < 1
The series diverges (no finite sum) when |r| ≥ 1.

Your Results

Calculated
Sum of the Series (Sₙ)
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Total of all n terms
Nth Term (aₙ)
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The last (nth) term in the series
Average Term
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Sₙ ÷ n
Formula Used
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Based on the series type selected

Ready

Choose a series type, enter the first term, difference or ratio, and number of terms, then press Calculate.

How the Sum of Series Calculator works

A series is the sum of the terms of a sequence. This calculator handles the two most common types with exact closed-form formulas: arithmetic series, where each term increases by a constant common difference d, and geometric series, where each term is multiplied by a constant common ratio r. Instead of adding up every term one by one, these formulas let you jump straight to the total.

Arithmetic series formula

For an arithmetic series with first term a₁, common difference d, and n terms, the nth term is aₙ = a₁ + (n − 1)d. The sum of the first n terms is Sₙ = n/2 × (2a₁ + (n − 1)d), which is the same as n times the average of the first and last term, Sₙ = n/2 × (a₁ + aₙ). For example, 1 + 2 + 3 + ... + 10 has a₁ = 1, d = 1, n = 10, giving a₁₀ = 10 and S₁₀ = 10/2 × (1 + 10) = 55.

Geometric series formula

For a geometric series with first term a₁, common ratio r, and n terms, the nth term is aₙ = a₁ × r^(n − 1). The sum of the first n terms is Sₙ = a₁ × (1 − rⁿ)/(1 − r) when r ≠ 1. If r = 1, every term equals a₁, so the sum simplifies to Sₙ = a₁ × n. For example, 3 + 6 + 12 + 24 + 48 has a₁ = 3, r = 2, n = 5, giving a₅ = 48 and S₅ = 3 × (1 − 2⁵)/(1 − 2) = 93. When |r| < 1 and n is allowed to grow without bound, the sum converges to the infinite-series total S = a₁/(1 − r).

Common sources of error

  • Mixing up d and r: arithmetic series use a common difference (added each step); geometric series use a common ratio (multiplied each step) — entering one where the other belongs gives a completely different total.
  • Off-by-one in n: the nth term is a₁ + (n − 1)d or a₁ × r^(n − 1), not a₁ + n·d or a₁ × rⁿ — the exponent/multiplier is always n − 1 because the first term itself is already "step zero."
  • Dividing by zero when r = 1: the geometric sum formula has (1 − r) in the denominator, so it is undefined at r = 1; use Sₙ = a₁ × n instead in that special case.

Checking your result

For a small arithmetic series, add the first and last term and multiply by n/2 by hand to confirm the calculator's output. For a small geometric series, add a handful of terms directly (a₁, a₁r, a₁r², ...) and compare the running total to Sₙ. If |r| > 1, the sum should grow quickly with n; if |r| < 1, later terms shrink toward zero and the sum should approach a limit.

Applications

Arithmetic series show up in evenly spaced schedules, stacking problems, and simple growth patterns (rows of seats, stacked pipes, equal pay raises each period). Geometric series appear in compound interest and loan amortization, population or investment growth models, and depreciation schedules — anywhere a quantity is repeatedly multiplied by a fixed factor.

Frequently Asked Questions

What is the formula for the sum of an arithmetic series?
The sum of the first n terms of an arithmetic series is Sₙ = n/2 × (2a₁ + (n − 1)d), where a₁ is the first term and d is the common difference. This is equivalent to Sₙ = n/2 × (a₁ + aₙ), the number of terms times the average of the first and last term.
What is the formula for the sum of a geometric series?
The sum of the first n terms of a geometric series is Sₙ = a₁ × (1 − rⁿ)/(1 − r), where a₁ is the first term and r is the common ratio (r ≠ 1). If r = 1, every term equals a₁, so the sum is simply Sₙ = a₁ × n.
What happens when the common ratio r equals 1 in a geometric series?
When r = 1 the standard formula Sₙ = a₁(1 − rⁿ)/(1 − r) divides by zero, so it cannot be used. Instead every term in the series equals a₁, so the sum of n terms is simply Sₙ = a₁ × n.
Can I find the sum of an infinite geometric series?
Yes, but only when the common ratio satisfies |r| < 1. In that case the terms shrink toward zero and the infinite sum converges to S = a₁/(1 − r). If |r| is greater than or equal to 1, the series does not converge and has no finite sum.