Formulas
aₙ = a₁ + (n − 1) · d
Sₙ = n / 2 · (2a₁ + (n − 1) · d) = n · (a₁ + aₙ) / 2
Where a₁ is the first term, d is the constant difference between consecutive terms, and n is the term index.
Worked example
Sequence 2, 5, 8, 11, ... with a₁ = 2 and d = 3. The 10th term:
- a₁₀ = 2 + (10 − 1) · 3 = 2 + 27 = 29
- S₁₀ = 10 / 2 · (2 + 29) = 5 · 31 = 155
- Mean = S₁₀ / 10 = 15.5
Note that the mean of an AP is exactly the average of the first and last terms.
FAQ
Can the common difference be negative?
Yes — a decreasing AP simply has d < 0, e.g., 100, 95, 90, 85, ... has d = −5. All formulas work the same.
What about partial sums?
Use the same Sₙ formula with smaller n. To find the sum from term i to term j: compute Sⱼ − S(i−1).