What is an arithmetic sequence?

An arithmetic sequence is a list of numbers where each term increases (or decreases) by a fixed amount called the common difference, $d$. Starting from a first term $a_1$, every following term adds $d$. This calculator finds the n-th term ($a_n$) and the sum of the first n terms ($S_n$) instantly from three inputs.

Number line showing equally spaced points forming an arithmetic sequence with a constant gap d — Each term increases by the same common difference $d$.

How to use this calculator

Enter the first term $a_1$, the common difference $d$ (positive for increasing sequences, negative for decreasing), and the term position $n$ you want to reach. Press calculate to see the value of $a_n$ and the cumulative sum $S_n$ of all terms from $a_1$ through $a_n$.

The formula explained

The n-th term is found with $$a_n = a_1 + (n - 1)d,$$ because you add the common difference $(n - 1)$ times after the first term. The partial sum uses the pairing trick discovered by Gauss: $$S_n = \frac{n}{2} \times (a_1 + a_n),$$ which is the average of the first and last term multiplied by how many terms there are.

Diagram pairing first and last terms of a sequence to illustrate the sum formula — Pairing terms from both ends gives the sum formula $S_n = \frac{n}{2}(a_1 + a_n)$.

Worked example

Suppose $a_1 = 2$, $d = 3$, and $n = 10$. The 10th term is $$a_n = 2 + (10 - 1)\cdot 3 = 2 + 27 = 29.$$ The sum of the first 10 terms is $$S_n = \frac{10}{2} \times (2 + 29) = 5 \times 31 = 155.$$

Comparing Arithmetic Sequences Across Scenarios

The two key outputs of an arithmetic sequence are the nth term $a_n = a_1 + (n-1)d$ and the partial sum $S_n = \frac{n}{2}(a_1 + a_n)$. The table below applies these formulas to several realistic input sets, including a positive common difference, a negative (decreasing) one, and a fractional step.

First term $a_1$	Common difference $d$	Terms $n$	nth term $a_n$	Sum $S_n$	Sequence preview
5	2	8	19	96	5, 7, 9, …, 19
10	-3	6	-5	15	10, 7, 4, …, -5
0	0.5	20	9.5	95	0, 0.5, 1, …, 9.5
100	-10	11	0	550	100, 90, 80, …, 0
1	1	100	100	5050	1, 2, 3, …, 100

Notice how a negative $d$ produces a decreasing sequence, and how the sum can still be positive even when later terms turn negative, as long as the early terms outweigh them.

Key Terms and Variables

First term $a_1$: The starting value of the sequence — the value at position $n = 1$. Every other term is built by repeatedly adding the common difference to it.
Common difference $d$: The fixed amount added from one term to the next: $d = a_{n} - a_{n-1}$. A positive $d$ gives an increasing sequence, a negative $d$ a decreasing one, and $d = 0$ a constant sequence.
nth term $a_n$: The value of the term at position $n$, found directly with $a_n = a_1 + (n-1)d$ without listing every term in between.
Partial sum $S_n$: The sum of the first $n$ terms, $S_n = \frac{n}{2}(a_1 + a_n)$. It pairs the first and last terms and multiplies by the number of pairs.
Term position $n$: A positive integer index telling you which term you want (1st, 2nd, 3rd, …). It also equals how many terms are being summed in $S_n$.
Arithmetic sequence vs. series: A sequence is the ordered list of terms (5, 7, 9, …); a series is what you get when you add those terms together. $a_n$ describes the sequence, while $S_n$ is the value of the corresponding finite series.

How to Calculate It by Hand

Use this procedure to find both the nth term and the sum from the three inputs $a_1$, $d$, and $n$. We'll carry the example $a_1 = 5$, $d = 2$, $n = 8$ through each step.

Identify $a_1$, $d$, and $n$. Read the first term, the constant step between terms, and the position you need. Here $a_1 = 5$, $d = 2$, and $n = 8$.
Compute the nth term. Substitute into $a_n = a_1 + (n-1)d$:
$a_8 = 5 + (8 - 1)\times 2 = 5 + 7\times 2 = 5 + 14 = 19$.
Compute the partial sum. Substitute $a_1$, $a_n$, and $n$ into $S_n = \frac{n}{2}(a_1 + a_n)$:
$S_8 = \frac{8}{2}(5 + 19) = 4 \times 24 = 96$.
Check the result. Listing the terms gives 5, 7, 9, 11, 13, 15, 17, 19 — the last is $a_8 = 19$ and they total 96, confirming $S_8$.

If you only need the sum and already prefer working from $a_1$ and $d$, the combined form $S_n = \frac{n}{2}\bigl(2a_1 + (n-1)d\bigr)$ gives the same answer in one line: $S_8 = \frac{8}{2}(2\times 5 + 7\times 2) = 4(10 + 14) = 96$.

FAQ

Can d be negative? Yes. A negative common difference produces a decreasing sequence, and the formulas still work exactly.

What does $S_n$ represent? It is the total of all terms from $a_1$ up to and including $a_n$ — a partial (finite) sum, not an infinite series.

What if $n = 1$? Then $a_n = a_1$ and $S_n = a_1$, since there is only one term.

Sum of first n terms (S_n)	96
First term (a₁)	5
Common difference (d)	2
Term position (n)	8

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