If the sum of n terms of an A.P. is and its mth term is 164,

Solution:

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 .