Arithmetic Nth Term Calculator
Enter the first term, common difference and term number to find the nth term and partial sum.
Result
How it is calculated
In an arithmetic sequence each term differs from the previous one by the same common difference d. The tool takes the first term a, the difference d and the position n, and returns an = a + (n − 1) × d and the sum of the first n terms, Sn = n / 2 × (2a + (n − 1) × d). Both are shown to 2 decimals, and the first term and difference are echoed back. a and d can be negative or decimal, so a falling sequence uses a negative d. The position n must be a whole number of at least 1; a value such as 2.5 or 0 gives no result.
Worked example
The sequence starts at 5 and goes up by 3 each step: 5, 8, 11, 14 and so on. For the 20th term, a20 = 5 + (20 − 1) × 3 = 5 + 57 = 62.00. The sum of the first 20 terms is S20 = 20 / 2 × (2 × 5 + 19 × 3) = 10 × (10 + 57) = 10 × 67 = 670.00. A quick check: the average of the first and last term is (5 + 62) / 2 = 33.5, and 33.5 × 20 = 670.
Common mistakes
- Using n as the number of steps instead of the position. The first term is n = 1, so the 20th term has only 19 differences added to the first.
- Forgetting the sign of d for a decreasing sequence. A sequence like 20, 17, 14 has d = −3, not 3.
- Entering the second term as the first term. If your list starts 8, 11, 14, then a = 8, not 5 or 11.
- Entering a non-integer position (such as 2.5) or zero. Positions are whole numbers starting from 1, so the tool gives no result for them.
- Applying the formula to a geometric sequence. Doubling or tripling each step is not a constant difference, so this tool does not fit.
About this calculator
An arithmetic sequence increases (or decreases) by the same fixed amount, the common difference (d), from one term to the next. Any term can be found directly without listing every value before it, using aₙ = a₁ + (n−1)d, where a₁ is the first term and n is the position you want.
The same sequence has a matching shortcut for adding up its first n terms: Sₙ = n/2 × (2a₁ + (n−1)d), avoiding tedious term-by-term addition. This is foundational algebra taught in middle and high school, and it comes up anywhere a quantity changes by a constant step each period — seating rows in a theater, stacked objects, or simple interest-style linear growth problems.
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Comments (1)
The tool goes from a position to a value, but you can turn it around. To find where a sequence reaches a target, use n = (target − a) ÷ d + 1; for first term 3 and difference 5, a target of 103 gives n = 21, and entering 21 confirms 103.00. With a negative difference the sum rises, then falls: 20, 17, 14 and so on peaks at 77.00 after 7 terms. Which sequence problems would you like handled?