If the sum of $n$ terms of an A.P. is $3 n^{2}+5 n$ and its $m^{\text {th }}$ term is 164, find the value of $m$.
Let a and b be the first term and the common difference of the A.P. respectively.
$a_{m}=a+(m-1) d=164 \ldots(1)$
Sum of $n$ terms, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
Here,
$\frac{n}{2}[2 a+n d-d]=3 n^{2}+5 n$
$\Rightarrow n a+\frac{d}{2} n^{2}-\frac{d}{2} n=3 n^{2}+5 n$
$\Rightarrow \frac{d}{2} n^{2}+\left(a-\frac{d}{2}\right) n=3 n^{2}+5 n$
Comparing the coefficient of $n^{2}$ on both sides, we obtain
$\frac{d}{2}=3$
$\Rightarrow d=6$
Comparing the coefficient of n on both sides, we obtain
$a-\frac{d}{2}=5$
$\Rightarrow a-3=5$
$\Rightarrow a=8$
Therefore, from (1), we obtain
$8+(m-1) 6=164$
$\Rightarrow(m-1) 6=164-8=156$
$\Rightarrow m-1=26$
$\Rightarrow m=27$
Thus, the value of $m$ is 27 .