# If the sum of n terms of an A.P. is 3n2 + 5n and its mth 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. = Sn = 3n2 + 5n ………….(I)

mth term of A.P. = tm = 164 ……………………….(III)

tm = 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

mth term = tm = 164 = 8 + (m – 1)6

164 – 8 = (m – 1)6

m – 1 = 156/6

m – 1 = 26

m = 26 + 1

m = 27

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