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

[Hint: t_m = S_m — S_m-1= 3m² + 5m — 3 (m— 1)² — 5 (m— 1) = 3 (2m — 1) + 5 = 6m + 2]

S_n = 3n² + 5n

Taking n = 1, we get

S₁ = 3(1)² + 5(1)

⇒ S₁ = 3 + 5

⇒ S₁ = 8

⇒ a₁ = 8

Taking n = 2, we get

S₂ = 3(2)² + 5(2)

⇒ S₂ = 12 + 10

⇒ S₂ = 22

∴ a₂ = S₂ – S₁ = 22 – 8 = 14

Taking n = 3, we get

S₃ = 3(3)² + 5(3)

⇒ S₃ = 27 + 15

⇒ S₃ = 42

∴ a₃ = S₃ – S₂ = 42 – 22 = 20

So, a = 8,

d = a₂ – a₁ = 14 – 8 = 6

Now, we have to find the value of m

a_n = a + (n – 1)d

⇒ a_m = 8 + (m – 1)6

⇒ 164 = 8 + 6m – 6

⇒ 164 = 2 + 6m

⇒ 162 = 6m

⇒ m = 27