Arithmetic Sequence Calculator
Work out any arithmetic sequence from three numbers: the first term a₁, the common difference d (negative is fine, decimals too), and how many terms n. You get the nth term, the sum of the first n terms, the explicit formula, and the sequence written out.
Example: with First term a₁ 3 · Common difference d 5 · Number of terms n 10 → nth term aₙ: 48 — term number 10.
- Sum of the first n terms Sₙ255 — total of terms 1 through 10
- Explicit formulaaₙ = 3 + (n − 1) × 5
- The sequence3, 8, 13, 18, 23, … , 48
Computed by the calculator below using its default values. Change any input to see your own numbers.
aₙ = a₁ + (n − 1)d and Sₙ = n(a₁ + aₙ)/2 — the pairing trick young Gauss reportedly used to sum 1 through 100 in seconds.
Equal steps make easy formulas
An arithmetic sequence climbs (or falls) by the same amount every step: 3, 8, 13, 18, … adds 5 each time. Because the growth is perfectly regular, you can jump straight to any term without listing the ones before it — the nth term is just the first term plus (n − 1) steps of size d: aₙ = a₁ + (n − 1)d. Ten terms into that sequence, a₁₀ = 3 + 9 × 5 = 48.
The sum has an equally clean shortcut. Pair the first term with the last, the second with the second-to-last, and so on: every pair totals a₁ + aₙ. With n terms there are n/2 pairs, so Sₙ = n(a₁ + aₙ)/2. It is the trick attributed to a young Gauss, who summed 1 through 100 as 50 pairs of 101 = 5,050.
Where these sequences show up
Anything that changes by a fixed amount per period is arithmetic: simple interest balances, seats per row in a theater that adds two seats each row, a loan principal shrinking by equal installments, mile markers on a highway. If the change is a fixed percentage instead of a fixed amount, you want a geometric sequence — that is compound growth, and its formulas look very different.
How it’s calculated
nth term: aₙ = a₁ + (n − 1)d. Sum of the first n terms: Sₙ = n(a₁ + aₙ)/2, equivalent to Sₙ = n[2a₁ + (n − 1)d]/2. n is rounded to the nearest whole number and must be at least 1; a₁ and d may be any real numbers. Results keep up to 6 decimal places.
Constant common difference by definition — if the gap between your data points drifts, the sequence is not arithmetic and these closed forms do not apply.
Classic arithmetic sequences (10 terms)
| Sequence | a₁ | d | a₁₀ | S₁₀ |
|---|---|---|---|---|
| Counting numbers | 1 | 1 | 10 | 55 |
| Odd numbers | 1 | 2 | 19 | 100 |
| Multiples of 5 | 5 | 5 | 50 | 275 |
| Countdown by tens | 100 | −10 | 10 | 550 |
Computed with aₙ = a₁ + (n − 1)d and Sₙ = n(a₁ + aₙ)/2.
Common mistakes
- Multiplying d by n instead of (n − 1): the first term has taken zero steps, so the 10th term uses 9 steps, not 10.
- Using these formulas on a geometric sequence (×2 each step is not +2 each step) — constant ratio needs aₙ = a₁rⁿ⁻¹ instead.
- Sign slips with negative d: a countdown like 100, 93, 86 has d = −7, and forgetting the minus turns descent into growth.
- Confusing aₙ (one term) with Sₙ (a running total) — a₁₀ = 48 is the tenth rung, S₁₀ = 255 is all ten rungs added up.
Frequently asked questions
What is the arithmetic sequence formula?
The nth term is aₙ = a₁ + (n − 1)d, where a₁ is the first term and d the common difference. The sum of the first n terms is Sₙ = n(a₁ + aₙ)/2.
How do I find the common difference?
Subtract any term from the one after it: d = a₂ − a₁. In 3, 8, 13 the difference is 5. If that gap is not the same everywhere, the sequence is not arithmetic.
Can the common difference be negative or a fraction?
Yes. d = −7 gives a decreasing sequence (100, 93, 86, …), and d = 0.5 climbs by halves (2, 2.5, 3, …). The formulas work unchanged for any real a₁ and d.
What is the difference between arithmetic and geometric sequences?
Arithmetic adds a constant (3, 8, 13 — plus 5); geometric multiplies by a constant (3, 6, 12 — times 2). Fixed dollar raises are arithmetic; percentage raises are geometric.
Why does the sum formula divide by 2?
Pairing terms from both ends: first + last, second + second-to-last — every pair adds to a₁ + aₙ, and n terms form n/2 such pairs. So Sₙ = n(a₁ + aₙ)/2, even when n is odd (the middle term counts as half a pair twice).