The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its A. 24th te

1.	The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its A. 24th term B. 27th term C. 26th termD. 25th term
Answer» 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.

The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its A. 24th term B. 27th term C. 26th termD. 25th term

Answer»

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.