Question

The sum of n terms of an A.P. is 3n² + 5n, then 164 is its 

A. 24^th term 

B. 27^th term 

C. 26^th term

D. 25^th term

Answer 1

S = 3n² + 5nS₁ 

= a₁ = 3 + 5 = 8S₂ 

= a₁ + a₂ = 12 + 10 = 22 

⇒ a₂ = S₂ – S₁ = 22 – 8 

= 14S₃ = a₁ + a₂ + a₃ 

= 27 + 15 = 42

⇒ a₃ = S₃ – S₂ 

= 42 – 22 = 20

∴ Given AP is 8,14,20,.....Thus a = 8, d = 6

Given t_m = 164.

164 = [a + (n –1)d]

164 = [(8) + (m –1)6]

164 = [8 + 6m – 6]

164 = [2 + 6m]

162= 6mm = 162/6.

∴ m = 27.