Q. 134.2( 113 Votes )

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

Answer :

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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