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13.  If the sum of n terms of an A.P. is  3 n^2 + 5 n  and its m^{th } term is 164, find the value of m.

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Given : If the sum of n terms of an A.P. is  3 n^2 + 5 n  and its m^{th } term is 164

Let a and d be first term and common difference of a AP ,respectively.

 Sum of n terms = 3 n^2 + 5 n

\Rightarrow \, \, \frac{n}{2}[2a+(n-1)d]=3n^2+5n

\Rightarrow \, \, 2a+(n-1)d=6n+10

\Rightarrow \, \, 2a+nd-d=6n+10

Comparing the coefficients of n on both side , we have

     \Rightarrow \, \, d=6

Also , 2a-d=10

\Rightarrow \, \, 2a-6=10

\Rightarrow \, \, 2a=10+6

\Rightarrow \, \, 2a=16

\Rightarrow \, \, a=8

 m th term is 164.

\Rightarrow \, \, a+(m-1)d=164

\Rightarrow \, \, 8+(m-1)6=164

\Rightarrow \, \, (m-1)6=156

\Rightarrow \, \, m-1=26

\Rightarrow \, \, m=26+1=27

Hence, the value of m is 27.

Posted by

seema garhwal

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