Question

If the sum of $n$ terms of an A.P. is $3n^2+5n$ 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, $S_n=3n^2+5n$
$\therefore S_m=3m^2+5m\text{ and }{S}_{m-1}=3(m-1{)}^2+5(m-1)$
Now, using the formula $T_m=S_m-S_{m-1}$, we have
$T_m=\left(3m^2+5m\right)-\left[3(m-1{)}^2+5(m-1)\right]$
$=\left(3m^2+5m\right)-\left[3\left(m^2+1-2m\right)+5m-5\right]$
$=\left(3m^2+5m\right)-\left[3m^2+3-6m+5m-5\right]$
$=3m^2+5m-3m^2-3+6m-5m+5=6m+2$
But given, $T_m=164$
$\therefore 6m+2=164$
$\Rightarrow 6m=164-2$
$\Rightarrow 6m=162$
$\Rightarrow m=\frac{162}{6}=27$