How this instrument works
An arithmetic sequence is a list built by repeating the same addition: start at a₁ and add the common difference d each step. Because every step adds exactly d, the nth term is the first term plus (n − 1) copies of that step: aₙ = a₁ + (n − 1)d. The subtraction of one inside the parentheses is easy to miss — it counts the steps taken to reach the nth term, not the count of terms passed, and it is the one spot where careless arithmetic on this formula usually breaks.
The sum formula rests on a trick a nine-year-old Carl Friedrich Gauss is said to have found for himself: pair the first term with the last, the second with the second-last, and so on — each pair adds to the same total, a₁ + aₙ. There are n ⁄ 2 such pairs, so Sₙ = n ⁄ 2 × (a₁ + aₙ), which expands to n ⁄ 2 × (2a₁ + (n − 1)d) once aₙ is substituted in. The pairing holds whether n is even or odd; with an odd count, the middle term pairs with itself and the algebra absorbs it without a special case.
Neither formula requires d to be positive. Set it to zero and every term equals a₁, so the sum collapses to n × a₁; set it negative and the sequence counts down, which is exactly how a straight-line depreciation schedule or a countdown behaves. The one thing both formulas assume is a fixed step size — once the gap between consecutive terms starts growing or shrinking, the list is no longer arithmetic and aₙ above will not describe it.
- Enter the starting value in First term (a₁) — the sequence's first number, before any steps are added.
- Enter the fixed step size in Common difference (d); use a negative value if the sequence counts down instead of up.
- Set Term number (n) to the position you want — n = 1 returns the first term itself, unchanged.
- Read nth term for aₙ, the value sitting at position n, and Sum of first n terms for the running total of every term from the first through the nth added together.
Worked example — starting at 2, stepping by 3
Take a₁ = 2, d = 3, and n = 5. The nth term is aₙ = 2 + (5 − 1) × 3 = 2 + 12 = 14. Counting the terms out directly confirms it: 2, 5, 8, 11, 14 — five terms, each three more than the last, landing on 14 at the fifth position.
The running total uses the same three inputs: Sₙ = 5 ⁄ 2 × (2 × 2 + (5 − 1) × 3) = 2.5 × (4 + 12) = 2.5 × 16 = 40. Adding the five terms by hand gives the identical check: 2 + 5 + 8 + 11 + 14 = 40, which is exactly what the pairing trick predicts — the pairs (2, 14) and (5, 11) each total 16, plus the unpaired middle term 8, giving 16 + 16 + 8 = 40.
Questions
How is the nth-term formula for an arithmetic sequence derived?
Starting from a₁, each step adds d once, so reaching the nth term takes exactly n − 1 additions of d — never n, because the first term already counts as the starting point before any step is taken. Counting steps this way gives aₙ = a₁ + (n − 1)d directly, with no other assumption needed.
Why does the sum formula divide by 2?
The division by 2 comes from Gauss's pairing trick: writing the sum forwards and backwards and adding the two copies term by term produces n pairs that each total (a₁ + aₙ), so twice the sum equals n × (a₁ + aₙ). Dividing by 2 undoes that doubling and leaves Sₙ = n ⁄ 2 × (a₁ + aₙ).
What happens if the common difference is zero or negative?
A zero common difference makes every term equal a₁, so the sequence is constant and Sₙ simplifies to n × a₁. A negative common difference produces a decreasing sequence — the nth-term and sum formulas need no change at all, since d is simply a signed number in both.
How is an arithmetic sequence different from a geometric one?
An arithmetic sequence adds a fixed amount each step, aₙ = a₁ + (n − 1)d; a geometric sequence multiplies by a fixed ratio instead. The practical sign is in the differences between consecutive terms: constant differences mean arithmetic, constant ratios mean geometric, and a sequence is rarely both unless it is also constant throughout.
Can the term number be a fraction or a non-integer?
Not meaningfully — n counts a specific term, so it should be a positive whole number, and the nth-term formula only has a clear reading at integer positions. The sum formula's algebra still computes for a non-integer n, but the result no longer corresponds to adding a whole number of terms.
What is the classic Gauss story behind this formula?
Gauss, as a schoolboy, is said to have summed 1 + 2 + … + 100 almost instantly by pairing 1 with 100, 2 with 99, and so on — fifty pairs each totaling 101, giving 5,050. That list is the arithmetic case a₁ = 1, d = 1, n = 100, and the pairing is exactly the derivation behind Sₙ = n ⁄ 2 × (a₁ + aₙ) above.