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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.

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seema garhwal

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