Question

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

• A. 27
• B. 25
• C. 30
• D. 32

Answer 1

Answer: (A)

Given, $S_n=3 n^2+5 n$
$\therefore S_m=3 m^2+5 m \text { and } S_{m-1}=3(m-1)^2+5(m-1)$
Now, using the formula $T_m=S_m-S_{m-1}$, we have
$T_m =\left(3 m^2+5 m\right)-\left[3(m-1)^2+5(m-1)\right] $
$ =\left(3 m^2+5 m\right)-\left[3\left(m^2+1-2 m\right)+5 m-5\right] $
$=\left(3 m^2+5 m\right)-\left[3 m^2+3-6 m+5 m-5\right] $
$ =3 m^2+5 m-3 m^2-3+6 m-5 m+5=6 m+2$
But given, $ T_m=164$
$\therefore 6 m+2 =164$
$\Rightarrow 6 m =164-2 $
$\Rightarrow 6 m =162 $
$\Rightarrow m =\frac{162}{6}=27$