Arithmetic Sequence Calculator

Background.

Ask for the 50th term of 3, 8, 13, 18, … and the quick answer is 3 + 49 × 5 = 248. Notice the 49. Fifty terms are joined by forty-nine gaps, and each gap adds the common difference once, which is why every arithmetic-sequence formula carries an (n − 1) where a first guess would put n. Use n instead and every answer lands one step too far.

An arithmetic sequence is any list that grows or shrinks by the same amount at every step. Four numbers describe a finite run of one completely: the first term a₁, the common difference d, the number of terms n and the last term aₙ. One equation ties them together, aₙ = a₁ + (n − 1)d, so any three of them fix the fourth. Choose which one you are missing and the calculator returns it along with the other three, the sum of all n terms, the explicit rule for that particular sequence and a preview of its opening terms.

The sum is often the more useful half. Rows of seats that each hold five more than the row in front, a loan balance that falls by the same slice of principal every month, the distance a dropped stone covers in each successive second: in each case the total over many steps matters more than any single step, and the total of an arithmetic run has a closed form, n(a₁ + aₙ) ÷ 2, that needs no adding up term by term.

Counting starts at 1. The first term is a₁, and n is always a whole number of at least 1. A few computer-science and discrete-mathematics texts start at a₀; if yours does, every position here reads one higher than in your book.

What is arithmetic sequence calculator?

An arithmetic sequence, or arithmetic progression, is a sequence in which the difference between any two consecutive terms is the same constant, the common difference d. A positive d makes the sequence rise, a negative d makes it fall, and d = 0 gives a constant sequence in which every term equals the first. Plotted against position, the terms sit on a straight line of slope d, which makes an arithmetic sequence the whole-number version of a linear function. A quick test follows: every term except the first and last is the average of its two neighbours, so 13 sits exactly midway between 8 and 18. The list of terms is the sequence; the total of the first n terms is the arithmetic series, Sₙ. Physics supplies a classic instance. A stone dropped from rest, ignoring air resistance, falls 4.9 m in the first second, 14.7 m in the second and 24.5 m in the third, a sequence with d = 9.8 m, and its five-second total of 5 × (4.9 + 44.1) ÷ 2 = 122.5 m is the familiar ½gt².

How to use this calculator.

1. 01Under 'What do you want to find?', choose the quantity you are missing: the nth term, the common difference, the first term or the number of terms.
2. 02Fill in the three fields that mode uses. The field being solved for is ignored, so whatever is already in it does no harm.
3. 03Enter the common difference with its sign, −7 for a sequence that falls by 7 at each step. Fractions and decimals such as 0.25 are fine.
4. 04When solving for the number of terms, enter the value you want to reach as aₙ. If it is not a member of the sequence you will be shown the fractional position it would occupy, which usually exposes a typo or a wrong sign.
5. 05Check the result against the preview line beneath it: the first six terms, the last term, the count and the direction confirm at a glance that you set up the sequence you meant.

The formula.

Walking from the first term to the nth takes n − 1 steps of size d, so aₙ = a₁ + (n − 1)d. The other three modes are that one equation rearranged. The common difference is d = (aₙ − a₁) ÷ (n − 1), which needs n of at least 2, since a single term has no gap to measure. The first term is a₁ = aₙ − (n − 1)d. The number of terms is n = (aₙ − a₁) ÷ d + 1, which needs a non-zero d and has a valid answer only when it comes out as a whole number of 1 or more; a value of aₙ that falls between two terms, or before the first, is not a member of the sequence and is reported as such rather than snapped to the nearest term.

The sum comes from writing the run twice, once forwards and once backwards, and adding the two rows column by column. Every column totals a₁ + aₙ, because as one row climbs by d the other falls by d. There are n columns and each term has been counted twice, so Sₙ = n(a₁ + aₙ) ÷ 2. Put another way, the sum is the number of terms times the average of the first and last terms. Substituting the nth-term formula gives the equivalent Sₙ = n[2a₁ + (n − 1)d] ÷ 2, handy when you know the step but not the last term.

That second form shows the sum is quadratic in n: over long runs, doubling the number of terms roughly quadruples the sum whenever d is not zero. It also means a total can be turned back into a term count by solving a quadratic, as one of the questions below shows. Sums that begin partway through a sequence follow by subtraction, since the terms from position m to position n add to Sₙ − Sₘ₋₁. Positions from 1 up to 10,000,000 are accepted.

A worked example.

Frequently asked questions.

References& sources.

1. [1]OpenStax (Rice University), College Algebra 2e, §9.2 'Arithmetic Sequences' — defines the common difference ('The difference between any two consecutive terms is a constant. This constant is called the common difference.') and gives the explicit formula aₙ = a₁ + d(n − 1). Retrieved 2026-07-29; open access.
2. [2]OpenStax (Rice University), College Algebra 2e, §9.4 'Series and Their Notations' — gives the arithmetic series formula Sₙ = n(a₁ + aₙ)/2. Retrieved 2026-07-29; open access.
3. [3]Encyclopedia of Mathematics (European Mathematical Society), 'Arithmetic progression' — independent check on both formulas: general term aₙ = a + (n − 1)d, sum Sₙ = (a₁ + aₙ)n/2, plus the characteristic property aₙ = (aₙ₊₁ + aₙ₋₁)/2. Agrees with OpenStax on the worked example (S₁₂ = 366). Retrieved 2026-07-29; open access.
4. [4]Wolfram MathWorld, 'Arithmetic Series' — the pairing argument that produces Sₙ = ½n(a₁ + aₙ), and the Gauss anecdote it is usually taught with. Retrieved 2026-07-29; open access.
5. [5]OpenStax (Rice University), Calculus Volume 1, §5.1 'Approximating Areas', rule 'Sums and Powers of Integers' — Σᵢ₌₁ⁿ i = n(n + 1)/2, the a₁ = 1, d = 1 special case, used as a check on this calculator. Retrieved 2026-07-29; open access.