Each question below is based on a number series.

Solution

ATQ, Given: Tₙ = a n² + b n + c Use the three given terms: For n = 2: T₂ = 4a + 2b + c = 13 …(i) For n = 3: T₃ = 9a + 3b + c = 24 …(ii) For n = 4: T₄ = 16a + 4b + c = 39 …(iii) Subtract (i) from (ii): (9a + 3b + c) − (4a + 2b + c) = 24 − 13 5a + b = 11 …(iv) Subtract (ii) from (iii): (16a + 4b + c) − (9a + 3b + c) = 39 − 24 7a + b = 15 …(v) Subtract (iv) from (v): (7a + b) − (5a + b) = 15 − 11 2a = 4 ⇒ a = 2 From (iv): 5a + b = 11 ⇒ 5×2 + b = 11 10 + b = 11 ⇒ b = 1 Now use (i) to find c: 4a + 2b + c = 13 4×2 + 2×1 + c = 13 8 + 2 + c = 13 ⇒ c = 3 So: Tₙ = 2n² + n + 3 Now find T₆: T₆ = 2×6² + 6 + 3 = 2×36 + 6 + 3 = 72 + 6 + 3 = 81 Answer = 81.