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Question

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

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Solution

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

am=a+(m1)d=164...(1)

Sum of n terms , Sn=n2[2a+(n1)d]

Here n2[2a+ndd]=3n2+5n

n(ad2)+n2d2=3n2+5n

Comparing the coefficient of n&n2 on both sides, we obtain

d2=3d=6&ad2a3=5a=8

Therefore, from (1) we obtain 8+(m1)6=164

(m1)6=1648=156m1=26m=27

Thus, the value of m is 27.

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