Find the nth term, sum of n terms, and list terms of arithmetic, geometric, Fibonacci, and custom sequences.
| nth term | aₙ = a₁ + (n−1)d |
| Sum of n terms | Sₙ = n/2 · (2a₁ + (n−1)d) |
| Or equivalently | Sₙ = n/2 · (a₁ + aₙ) |
| First term (a₁) | 2 |
| Common difference (d) | 3 |
| Term 10 (aₙ) | 29 |
| Sum of 10 terms | 155 |
| Name | Pattern | First terms |
|---|---|---|
| Arithmetic | a, a+d, a+2d, … | 2, 5, 8, 11, … |
| Geometric | a, ar, ar², … | 3, 6, 12, 24, … |
| Fibonacci | F(n−1)+F(n−2) | 1, 1, 2, 3, 5, 8, … |
| Square numbers | n² | 1, 4, 9, 16, 25, … |
| Cube numbers | n³ | 1, 8, 27, 64, 125, … |
| Triangular | n(n+1)/2 | 1, 3, 6, 10, 15, … |
| Prime numbers | — | 2, 3, 5, 7, 11, 13, … |
A schedule starts at 2 and climbs by 3 each step, and someone needs the 10th entry. Writing a1 + n × d gives 32; the right term is a1 + (n − 1) × d = 29, because the first term has not had a step added yet. Carry the slip into the sum and the 10-term total goes from 155 to 170. This calculator uses the (n − 1) form, so the first term is exactly a1, and keeps the same bookkeeping for geometric and Fibonacci sequences.
Start with the defaults: first term 2, common difference 3, nth term at n = 10, sum of the first 10 terms. The listed sequence reads 2, 5, 8, 11, 14, 17, 20, 23, 26, 29.
The nth term formula is aₙ = a₁ + (n − 1) × d. At n = 10 that is 2 + 9 × 3 = 29, which matches the last entry in the list.
For the sum, the page uses Sₙ = n ÷ 2 × (2a₁ + (n − 1) × d), which here is 5 × (4 + 27) = 155. The same answer comes from averaging the first and last terms, (2 + 29) ÷ 2 = 15.5, and multiplying by 10 terms.
Geometric sequences multiply instead of add: aₙ = a₁ × r^(n−1) and Sₙ = a₁ × (1 − rⁿ) ÷ (1 − r). With first term 2 and ratio 3, term 10 is 2 × 3⁹ = 39,366 and the first ten terms add to 59,048. When the ratio is between -1 and 1, an infinite sum a₁ ÷ (1 − r) is shown as well: for a₁ = 1 and r = 0.5, ten terms give 1.998046875 and the infinite sum is 2.
Fibonacci starts 1, 1 and adds the previous two terms, so F(10) = 55 and the first ten add to 143. Displayed values keep 12 significant digits, so F(70) = 190,392,490,709,135 appears as 190,392,490,709,000.
So you can compare different points, such as the value at term 12 against the running total through term 20, without retyping everything.
It reports an unknown type, because the differences (3, 5, 7) and ratios (4, 2.25, 1.78) are not constant. It does not detect squares or predict a next term. Any list of exactly two numbers is labelled arithmetic.
No. For a ratio of -1 the page applies the ratio-of-1 shortcut and returns a₁ × n, while the true sum alternates between a₁ and 0. Work that case out by hand.
It starts 1, 1, so F(1) = 1. Texts that start 0, 1, 1 give the same values for every term from F(1) onward, so F(10) = 55 either way.