If the sum of n terms of an A.P. is 3n² + 5n and its m^th term is 164, find the value of m.

Let a and d be the first term and common difference of A.P.

Given, Sum of n terms of A.P. = S_n = 3n² + 5n ………….(I)

m^th term of A.P. = t_m = 164 ……………………….(III)

⇒ t_m = a + (m – 1)d …………………..(IV)

Equating (I) and (II)

⇒ 2an + n²d – dn = 6n² + 5n

⇒ (2a – d)n + n²d = 6n² + 10n

Comparing the coefficients of n², we get

d = 6

Comparing the coefficients of n, we get

2a – d = 10

Putting the value d = 6

⇒ 2a – 6 = 10 ⇒ 2a = 10 + 6

⇒ 2a = 16

⇒ a = 8

m^th term = t_m = 164 = 8 + (m – 1)6

⇒ 164 – 8 = (m – 1)6

⇒ m – 1 = 156/6

⇒ m – 1 = 26

⇒ m = 26 + 1

∴ m = 27