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

S = 3n2 + 5n
S1 = a1 = 3 + 5 = 8
S2 = a1 + a2 = 12 + 10 = 22
 ⇒ a2 = S2 – S1 = 22 – 8 = 14
S3 = a1 + a2 + a3 = 27 + 15 = 42
⇒ a3 = S3 – S2 = 42 – 22 = 20
∴ Given AP is 8,14,20,.....
Thus a = 8, d = 6
Given tm = 164.
164 = [a + (n –1)d]
164 = [(8) + (m –1)6]
164 = [8 + 6m – 6]
164 = [2 + 6m]
162= 6m
m = 162 / 6.
∴ m = 27.

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