Arithmetic Series Sum

Series Setup

S₁₀₀ = 100/2 · (2·1 + 99·1)

First term (a₁)

Common diff (d)

Number of terms (n)

Uses Sₙ = n/2·(2a₁ + (n−1)d). Also computes aₙ = a₁ + (n−1)d.

Examples

S₅₀ = 50/2 · (1 + 99)

First term (a₁)

Last term (aₙ)

Number of terms (n)

Uses Sₙ = n/2·(a₁ + aₙ). Simpler when you already know the last term.

Examples

Result

Enter values above and press Calculate Sum to see Sₙ, the series written out, and the average term.

—

Sum Sₙ

—

Last term aₙ

—

Average (a₁+aₙ)/2

Two Equivalent Formulas

Sₙ = n/2 · (a₁ + aₙ) Sₙ = n/2 · (2a₁ + (n−1)d)

Average formula: Sₙ = n/2·(a₁ + aₙ) pairs the first and last terms. Their average is (a₁ + aₙ)/2 and there are n terms, so multiplying gives the total sum.

Common-difference formula: Sₙ = n/2·(2a₁ + (n−1)d) works when you know d directly. It substitutes aₙ = a₁ + (n−1)d into the first formula.

Both formulas always give the same answer — choose whichever fits the information you have.

The average of any arithmetic sequence equals (a₁ + aₙ)/2. Multiplying by n gives the sum in one step.

Gauss's Trick

50 pairs × 101 = 5,050

Legend has it that a young Carl Friedrich Gauss — around age 9 — was asked to add all integers from 1 to 100. While classmates worked term by term, Gauss noticed a pattern.

Pair the first with the last: 1 + 100 = 101. Pair the second with the second-to-last: 2 + 99 = 101. Every pair sums to 101.

With 100 terms, there are 50 such pairs. So the total is 50 × 101 = 5,050.

This is exactly what Sₙ = n/2·(a₁ + aₙ) computes: n/2 = 50 pairs, each worth a₁ + aₙ = 101.

The formula works for any n — even odd values of n.
Works with negative terms, fractions, and decimals.
If d = 0, every term equals a₁ and Sₙ = n · a₁.
n = 1 is a valid edge case: S₁ = a₁.

Series and sequences giving you trouble?