A sequence {Tₙ} is defined by: Tₙ = a·n² + b·n + c for all positive integers n, where a, b, c are constants. It is known that - T₂ = 11, T₄ = 27, T₅ = 38 What is the value of T₆?

Given Tₙ = a n² + b n + c. Use the three given terms: For n = 2: T₂ = 4a + 2b + c = 11 …(1) For n = 4: T₄ = 16a + 4b + c = 27 …(2) For n = 5: T₅ = 25a + 5b + c = 38 …(3) Subtract (1) from (2): (16a + 4b + c) − (4a + 2b + c) = 27 − 11 12a + 2b = 16 6a + b = 8 …(A) Subtract (2) from (3): (25a + 5b + c) − (16a + 4b + c) = 38 − 27 9a + b = 11 …(B) Subtract (A) from (B): (9a + b) − (6a + b) = 11 − 8 3a = 3 a = 1 From (A): 6(1) + b = 8 ⇒ b = 2 From (1): 4a + 2b + c = 11 4(1) + 2(2) + c = 11 4 + 4 + c = 11 c = 3 So: Tₙ = n² + 2n + 3 Now: T₆ = 6² + 2×6 + 3 = 36 + 12 + 3 = 51 So T₆ = 51.