In an arithmetic progression, the 5th term is three times the 2nd term and the 10th term is 57. What is the sum of the first 15 terms?

Let first term a and common difference d.

a + 4d = 3(a + d) → a + 4d = 3a + 3d → d = 2a.

a + 9d = 57 → a + 18a = 57 → a = 3, d = 6.

S₁₅ = (15/2)[2a + 14d] = (15/2)(6 + 84) = (15/2)(90) = 675.

Check: terms 3, 9, 15, 21, 27 … the 5th (27) is 3 × the 2nd (9) ✓ and the 10th is 3 + 54 = 57 ✓.

Tip: S₁₅ = 15 × (8th term) = 15 × 45 = 675 — for an odd number of terms, the sum is n times the middle term.