A sequence has six terms: 6, 15, 28, ?, 66,

Solution

Given Tₙ = a n² + b n + c and the second difference is constant 4. For a quadratic sequence, the constant second difference = 2a. So 2a = 4 ⇒ a = 2. Thus Tₙ = 2n² + b n + c. Use the known terms. For n = 1: T₁ = 6 = 2·1² + b·1 + c = 2 + b + c ⇒ b + c = 4 …(1) For n = 2: T₂ = 15 = 2·2² + 2b + c = 8 + 2b + c ⇒ 2b + c = 7 …(2) Subtract (1) from (2): (2b + c) − (b + c) = 7 − 4 b = 3 From (1): b + c = 4 ⇒ 3 + c = 4 ⇒ c = 1 So Tₙ = 2n² + 3n + 1. Now find the 4th term: T₄ = 2·4² + 3·4 + 1 = 2·16 + 12 + 1 = 32 + 12 + 1 = 45 So the missing term is 45.