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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 $m$
$27$
$27$
$27$
$27$
Solution
Let $a$ and $b$ be the first term and the common difference of the $A.P.$ respectively.
$a_{m}=a+(m-1) d=164$ …………$(1)$
Sum of $n$ terms: $S_{n}=\frac{n}{2}[2 a+(n-1) d]$
Here,
$\frac{n}{2}[2 a+n d-d]=3 n^{2}+5 n$
$\Rightarrow n a+n^{2} \cdot \frac{d}{2}-\frac{n d}{2}=3 n^{2}+5 n$
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 $27 .$
