Arithmetic Series Calculator
An arithmetic progression (AP) is a sequence where each term differs from the previous by a constant d. Enter the first term a, common difference d and number of terms n to instantly get the nth term aₙ, the sum Sₙ of the first n terms, the mean and a preview of the first 30 terms. Widely used in school maths (GCSE / A-Level / HKDSE), linear depreciation, and savings ladders.
First term and common difference must be finite. Number of terms must be an integer between 1 and 10,000.
nth term aₙ
100
Sum of first n terms Sₙ
5,050
Mean a̅
50.5
First 30 terms
At most the first 30 terms are listed — extend with the formula.
aₙ = a + (n − 1) · d Sₙ = n/2 · (2a + (n − 1) · d) = n · (a + aₙ) / 2
aₙ = a + (n − 1) · d Sₙ = n/2 · (2a + (n − 1) · d)
Each term differs from the previous by a fixed value d. d > 0 → increasing, d < 0 → decreasing, d = 0 → constant.
Formula
aₙ = a + (n − 1) · d Sₙ = n/2 · (2a + (n − 1) · d) = n · (a + aₙ) / 2 Mean a̅ = Sₙ / n = (a + aₙ) / 2
- · First term a and common difference d may be any real numbers; this tool supports up to 10,000 terms.
- · When d = 0 the sequence is constant; Sₙ = n · a.
- · When d > 0 the sequence increases; when d < 0 it decreases. The sum formula works for both.
- · Sₙ = n/2 · (first term + last term) — equivalent to pairing terms from the ends. This is how Gauss famously summed 1 + 2 + … + 100 = 5050.
- · Arithmetic mean: if a, b, c form an arithmetic progression, then b = (a + c) / 2.
- · A geometric series (constant ratio rather than constant difference) needs the dedicated calculator on this site.
- · References: Stewart, Calculus (sequences appendix); standard A-Level / SAT II / IB Mathematics sequence chapters.
Frequently asked
How is an arithmetic series different from a geometric series?
In an arithmetic progression each term adds a fixed amount d (e.g. 2, 5, 8, 11 with d = +3). In a geometric progression each term multiplies by a fixed ratio r (e.g. 2, 6, 18, 54 with r = 3). Arithmetic captures linear growth (fixed monthly raise, equal instalments); geometric captures exponential growth (compound interest, population doubling). Sum formulas differ: Sₙ = n/2 · (2a + (n − 1)d) for arithmetic; Sₙ = a · (1 − rⁿ) / (1 − r) for geometric.
Why does Sₙ equal n/2 · (first term + last term)?
Gauss as a child summed 1 + 2 + … + 100 by pairing terms from the ends: (1 + 100) + (2 + 99) + … gives 50 pairs of 101, so 50 · 101 = 5050. In general: write Sₙ once forwards and once backwards, add the two — each pair adds to (a + aₙ) and there are n pairs, giving 2Sₙ = n · (a + aₙ), so Sₙ = n · (a + aₙ) / 2. Substituting aₙ = a + (n − 1)d gives the standard form Sₙ = n/2 · (2a + (n − 1)d).
How do I find the number of terms n given only the first term, last term and the sum?
Rearranging Sₙ = n · (a + aₙ) / 2 gives n = 2 · Sₙ / (a + aₙ). Example: a = 1, aₙ = 100, Sₙ = 5050 → n = 2 · 5050 / 101 = 100. If instead you know a, d and aₙ, use n = (aₙ − a) / d + 1.
Can I use this for a monthly savings ladder?
Yes, as long as the monthly contribution increases by a constant amount (linear growth). Example: start at $1,000 in month 1 and add $200 each month; in month 12 you contribute $3,200. With a = 1000, d = 200, n = 12 the total is Sₙ = 12 · (1,000 + 3,200) / 2 = $25,200. If instead contributions grow by a fixed percentage each month (compounding), use the geometric / compound interest calculator.
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