Sum of an Arithmetic Series

The closed form for the sum of an arithmetic series lets you skip adding term by term. The formula is S_n = n/2 (a₁ + a_n) or equivalently S_n = n/2 [2a₁ + (n-1)d]. It works because of the pairing trick attributed to a young Gauss: pair the first term with the last, the second with the second-last, and so on. Each pair sums to the same total (a₁ + a_n), and there are n/2 pairs. If n is odd, the middle term stands alone but the formula still holds. OpenStax Precalculus 2e section 11.4 gives this as the standard arithmetic series formula.

To use the arithmetic series formula you need four numbers: the first term a₁, the last term a_n, the number of terms n, and the common difference d. When you know three of them, solve for the fourth. For example, given a₁ = 3, d = 4, and S_n = 253, substitute into S_n = n/2 [2a₁ + (n-1)d] to get 253 = n/2 [6 + (n-1)4]. That simplifies to a quadratic: 2n² + n, 253 = 0. The positive solution is n = 11 (OpenStax Precalculus 2e example). To find a₁ given d = 5, n = 12, S_n = 402, use the same formula: 402 = 12/2 [2a₁ + 55] → 67 = a₁ + 27.5 → a₁ = 21. Every variable in the arithmetic series formula is recoverable by algebra.

Number of Terms Mistake

The most common error is counting n as (upper bound, lower bound) instead of (upper bound, lower bound + 1). For the series 1, 4, 7, 10, the index runs 1 to 4, so n = 4, not 3. Check your count before plugging into the closed form.

Arithmetic series sigma notation uses the Greek capital Σ to compactly write the sum. The three-part structure is: index variable, lower bound, upper bound, and summand. For the arithmetic series 1 + 4 + 7 + ... + 58, the first term a₁ = 1 and common difference d = 3. The n^th term formula is a_n = 1 + (n-1)3, so the series from term 1 to term 20 is written Σ_i=1²⁰ (1 + (i-1)3). The index variable i is a dummy variable; renaming it to j or k does not change the sum. The lower bound is the starting index, not the first term value. Evaluate the summand at each integer from the lower bound to the upper bound inclusive. For the sum of consecutive integers 1 + 2 + ... + 100, write Σ_i=1¹⁰⁰ i, and the closed form is n(n+1)/2. The arithmetic series sigma notation directly matches the S_n = n/2 (a₁ + a_n) formula when you identify a₁ and a_n from the summand.

Index Variable Is a Dummy

Σ_i=1ⁿ a_i and Σ_j=1ⁿ a_j are identical sums. The index letter only marks where the changing integer goes. Do not let a different letter confuse you when you see an arithmetic series sigma notation in a textbook or on a test.

Example 1: First 20 terms of 1, 4, 7, 10, ...
a₁ = 1, d = 3, n = 20. a₂₀ = 1 + (19)(3) = 58. S₂₀ = 20/2 (1 + 58) = 10 × 59 = 590. OpenStax Precalculus 2e section 11.4 gives this exact result.

Example 2: First 50 terms of 2, 5, 8, 11, ...
a₁ = 2, d = 3, n = 50. a₅₀ = 2 + (49)(3) = 149. S₅₀ = 25 (2 + 149) = 25 × 151 = 3775 (OpenStax).

Example 3: Find n when a₁ = 5, d = 3, S_n = 440.
440 = n/2 [10 + (n-1)3] → 880 = n(3n + 7) → 3n² + 7n − 880 = 0.Check: S₁₆ = 16/2 [10 + 45] = 8 × 55 = 440 (OpenStax).

Example 4: Find a₁ when d = 4, n = 15, S_n = 495.
495 = 15/2 [2a₁ + (14)(4)] → 495 = 7.5(2a₁ + 56) → 66 = 2a₁ + 56 → 10 = 2a₁ → a₁ = 5. OpenStax Precalculus 2e has a₁ = 19 for different inputs; here the arithmetic yields 5.

Derivation Diagram

Write the series forward: a₁, a₁+d, a₁+2d, ..., a_n. Write it backward: a_n, a_n-d, a_n-2d, ..., a₁. Add the two rows termwise. Each column sums to a₁+a_n. There are n columns, so 2S_n = n(a₁+a_n). Divide by 2 to get the arithmetic series formula. This pairing works for any n.

Off-by-one error in n. For Σ_i=3⁷ i, the number of terms is 7-3 + 1 = 5, not 4. Always add 1 after subtracting bounds.

Confusing Σx² with (Σx)². Σ_i=1ⁿ x_i² means square each term then add. (Σ_i=1ⁿ x_i)² means add then square. They are not equal except for trivial lists.

Using the wrong closed form. The arithmetic series formula is for constant difference between terms. Do not apply it to a geometric series, which uses a constant ratio and has its own formula S_n = a(1-rⁿ)/(1-r).

Misreading the lower bound as the first term value. In Σ_i=5¹⁰ (2i+1), the index starts at 5, so the first term is 2(5)+1 = 11, not 2(1)+1 = 3.

Forgetting parentheses in calculator entry. Entering Σ_i=1⁵ i+1 without parentheses around the summand gives (i)+1, not (i+1). The correct entry is Σ (i+1) from 1 to 5.

Variable	Definition	Where to Get It
a₁	First term of the series	Given in problem or list; e.g., sequence 1,4,7 gives a₁=1
aₙ	Last term (nth term)	Compute with a₁ + (n-1)d or read from the last listed term
n	Number of terms	Count terms in list, or use (upper bound – lower bound + 1) in sigma notation
d	Common difference	Subtract any term from the next; e.g., sequence 2,5,8 gives d=3
Sₙ	Sum of first n terms	Closed form gives this directly; verify with term-by-term addition for small n

The arithmetic series formula works only when the terms have a constant difference. If the sequence is not arithmetic, the formula gives a wrong number. Before applying S_n = n/2 (a₁ + a_n), confirm that the difference between successive terms is the same throughout. That single check saves you from the most common failure: applying a closed form to a series that does not have one.