Arithmetic Sequence: nth Term and Sum

From the first term and the common difference, finds the nth term a + (n−1)d of an arithmetic sequence and the sum of the first n terms.

Explanation

In an arithmetic sequence, every term sits a fixed distance from the one before it. That distance is the common difference. Given the first term, the difference and a term number, you can get both the nth term and the running total up to it.

a_n = a + (n-1)d

S_n = \dfrac{n(a + a_n)}{2}

$a$ — the first term
$d$ — the common difference, added at each step
$n$ — which term you want, an integer of 1 or more
$a_n$ — the nth term
$S_n$ — the sum of the first $n$ terms

Note the $n - 1$ . Reaching the nth term takes $n - 1$ steps, not $n$ , because you start out already standing on the first term. That off-by-one is the classic trap.

Example

Start with $a = 3$ , $d = 4$ and $n = 10$ . The sequence runs 3, 7, 11, 15, and so on.

a_{10} = 3 + (10 - 1) \times 4 = 3 + 36 = 39

S_{10} = \dfrac{10 \times (3 + 39)}{2} = \dfrac{420}{2} = 210

The tenth term is 39, and the first ten terms add to 210.

Notes

$n$ has to be a whole number of 1 or more. Fractions and non-positive values are rejected.
The common difference may be negative, in which case the terms shrink as the sequence goes on.
Read the sum formula as (first term + last term) times the number of terms, all halved. You are averaging the two ends and multiplying by how many terms there are.
Why it works: write the sequence forwards, write it again backwards underneath, and add column by column. Every column totals first term plus last term, and there are $n$ columns.