04 // Interactive Formula Compendium // CALCULUS-ALGEBRAS_n = (n / 2) · [ 2a₁ + (n − 1)d ] = (n / 2) · (a₁ + a_n)
Arithmetic Sequence & Series Sum Formula — Gauss Pairing Sheet
An arithmetic series is the sum of n terms with constant difference d between consecutive entries; its closed-form total is S_n = (n/2)(a₁ + a_n) = (n/2)[2a₁ + (n − 1)d].
Formula Reference
Main Formula
S_n = (n / 2) · [ 2a₁ + (n − 1)d ] = (n / 2) · (a₁ + a_n)
Rearranged Forms
a_n = a₁ + (n − 1)d (n-th Term Formula)
d = a_{k+1} − a_k (Common Difference)
Σ_{k=1..n} k = n(n + 1) / 2 (Sum of First n Natural Numbers)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| S_n | Sum of First n Terms (S_n) | dimensionless |
| a1 | First Term (a₁) | 1 |
| d | Common Difference (d) | 1 |
| n | Number of Terms (n) | integer |
Inputs & Parameters
[1]
[1]
[integer]
Calculated Result (Sum of First n Terms (S_n))
S_n =5,050dimensionless
Formula Notes & Background
- Write S_n forward: S_n = a₁ + (a₁ + d) + … + a_n, and backward: S_n = a_n + (a_n − d) + … + a₁.
- Add the two equations term-by-term: 2S_n = (a₁ + a_n) + (a₁ + a_n) + … n times = n(a₁ + a_n).
- Substitute a_n = a₁ + (n − 1)d and divide by 2 to get S_n = (n/2)[2a₁ + (n − 1)d].
Step-by-Step Examples
#1: Find the sum of the integers from 1 to 100 (Gauss's classic classroom problem).
a₁ = 1d = 1n = 100
S₁₀₀ = (100 / 2) × [2(1) + (100 − 1)(1)] = 50 × 101
→ 5,050
#2: Sum the first 20 terms of the arithmetic progression 5, 9, 13, 17, …
a₁ = 5d = 4n = 20
S₂₀ = (20 / 2) × [2(5) + 19(4)] = 10 × (10 + 76)
→ 860 (with 20th term a₂₀ = 81)
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Frequently Asked Questions
How do you find the number of terms n when given a₁, a_n, and d?+
Rearrange the n-th term formula a_n = a₁ + (n − 1)d to get n = ((a_n − a₁) / d) + 1.
What is the average value of all terms in an arithmetic progression?+
Because the spacing is symmetric, the arithmetic mean (and median) of the entire sequence is simply the average of the first and last terms: (a₁ + a_n) / 2.
Does an infinite arithmetic series ever converge?+
No, unless both a₁ = 0 and d = 0, the partial sums S_n grow without bound as O(n²).