Arithmetic Series Summation Calculator
An arithmetic series sums the terms of a sequence that increases (or decreases) by a constant common difference d: Sₙ = n/2 × (2a₁ + (n−1)d), equivalent to n times the average of the first and last term. Enter the first term, the number of terms, and the common difference to get the total sum and the last term.
The Pairing Trick Behind the Formula
Write the sum forwards and backwards, then add the two versions term by term: each pair (first+last, second+second-to-last, ...) adds up to the same total, a₁+aₙ. With n such pairs counted twice, the sum is n×(a₁+aₙ)/2 — substitute aₙ = a₁+(n−1)d and you get Sₙ = n/2×(2a₁+(n−1)d).
Worked Example
Take a₁=3, d=5, n=5, the sequence 3, 8, 13, 18, 23. The last term is aₙ = 3+4×5 = 23, and the sum is S₅ = 5/2×(6+20) = 65 — check it directly: 3+8+13+18+23 = 65.
Geometric Series Need a Different Formula
If your sequence multiplies by a constant ratio instead of adding a constant difference — 2, 6, 18, 54 rather than 3, 8, 13, 18 — it's geometric, not arithmetic, and needs Sₙ = a₁(1−rⁿ)/(1−r) instead. The sequence solver calculator on this site handles both types side by side.
Frequently Asked Questions
What does this calculator sum — any function or just arithmetic sequences?
Just arithmetic sequences — terms that increase or decrease by the same fixed amount each step. Summing an arbitrary function's values (sigma notation of f(i) for a general f) needs a different, case-by-case tool; this one solves the specific closed-form Σ(a₁+(i−1)d) for i=1 to n.
Where does the formula Sₙ = n/2(2a₁+(n−1)d) come from?
Pair the first term with the last, the second with the second-to-last, and so on — each pair sums to the same value, a₁+aₙ. There are n/2 such pairs, so the total is n/2 times (a₁+aₙ), which expands to n/2×(2a₁+(n−1)d) once you substitute aₙ = a₁+(n−1)d.
What if the common difference is negative?
The sequence decreases instead of increasing, and the formula still works unchanged — a negative d just makes later terms smaller, and the sum can come out negative too if the sequence crosses zero.
Can I use this for a sequence that isn't arithmetic?
No — if consecutive terms don't share a constant difference (for example a geometric sequence, which has a constant ratio instead), this formula gives the wrong answer. Check that your terms differ by the same fixed amount before trusting the result.
What's the sum of the first n positive integers?
That's the special case a₁=1, d=1: Sₙ = n/2×(2+(n−1)) = n(n+1)/2. For n=10, that's 10×11/2=55 — the same answer Gauss is famously said to have found as a schoolboy.