Arithmetic Series Sum (First n Terms)
Sum the first n terms of an arithmetic sequence from its first term and common difference — or from the first and last terms, or any two terms — sum only terms m to n, or find how many terms reach a target sum, with the nth-term and sum formulas, a term table and full worked steps.
How it works
aₙ = a₁ + (n − 1)d · Sₙ = n ÷ 2 × (a₁ + aₙ) = n ÷ 2 × [2a₁ + (n − 1)d] · d = (aₙ − a₁) ÷ (n − 1) = (a_q − aₚ) ÷ (q − p) · terms m to n = Sₙ − Sₘ₋₁ · terms for a sum S: (d/2)n² + (a₁ − d/2)n − S = 0
An arithmetic series is the running total of terms that each change by the same step d. Rather than adding term by term, pair the first and last terms: their average is the average of the whole sequence, so the sum is that average times the number of terms. This is the trick Gauss is said to have used to add 1 through 100 in seconds. Expanding aₙ gives the second form, Sₙ = n/2 × [2a₁ + (n − 1)d], which needs only the first term and the difference. You do not always start from a₁ and d. Given the first and last terms, the difference is the gap divided by n − 1; given any two terms aₚ and a_q, it is their difference divided by q − p, and the first term follows. To add only a block of terms m to n, subtract the sum of the first m − 1 terms. And because Sₙ is a quadratic in n, you can solve for how many terms it takes to reach a target sum. The calculator does all of these, writes the nth-term and sum formulas for your sequence, lists the terms with running totals and shows every step with your numbers.
Worked example
For a₁ = 3, d = 4, n = 10: the last term is 3 + 9 × 4 = 39, and the sum is 10 ÷ 2 × (3 + 39) = 5 × 42 = 210. The same sequence is aₙ = 4n − 1 with Sₙ = 2n² + n. Adding only terms 6 to 10 gives 210 − S₅ = 210 − 55 = 155. Working backwards, a sum of 210 needs 2n² + n − 210 = 0, so n = 10 terms.
Sources: Arithmetic progression — Wikipedia
Frequently asked questions
What is the formula for the sum of an arithmetic series?
Sₙ = n/2 × (first term + last term), where the last term is a₁ + (n − 1)d. Substituting gives Sₙ = n/2 × [2a₁ + (n − 1)d]. It works because the average of the first and last terms equals the average of every term in the sequence.
How do I find the sum if I only know two terms?
Find the common difference first: d = (a_q − aₚ) ÷ (q − p). Then the first term is a₁ = aₚ − (p − 1)d, and the usual sum formula applies. For example, a₂ = 7 and a₅ = 19 give d = 12 ÷ 3 = 4 and a₁ = 3.
How do I add only some of the terms, say terms 6 to 10?
Subtract the sum of the terms you are skipping: S₁₀ − S₅. For 3, 7, 11, … that is 210 − 55 = 155. Set “Start the sum at term” under More options and the calculator does it.
How many terms do I need to reach a given sum?
Set Sₙ equal to the target and solve the quadratic (d/2)n² + (a₁ − d/2)n − S = 0 for the smallest positive n. If the root is not a whole number, round up to get the first n whose sum passes the target. Choose “Target sum” to have it solved.
Can the common difference be negative?
Yes. A negative d gives a decreasing sequence, and the same formulas apply. For example a₁ = 20, d = −3, n = 5 sums to 5/2 × (20 + 8) = 70. A decreasing sequence may never reach a large target sum — the calculator tells you when that happens.
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