Question

Find the $n^{th}$ term of an A.P., sum of whose terms is $2n+3n^2$.

Answer 1

$S_n=2n+3n^2.....\left(1\right)\left[\text{Given}\right]$

$\therefore S_{n-1}=2(n-1)+3{(n-1)}^2$

$\Rightarrow S_{n-1}=2n-2+3n^2+3-6n$

$\Rightarrow S_{n-1}=1-6n+2n-3n^2$

$\Rightarrow S_{n-1}=1-6n+S_n\left[\text{From}{eq}^n\left(1\right)\right]$

$\Rightarrow S_n-S_{n-1}=6n-1$

$\Rightarrow T_n=6n-1[\because T_n=S_n-S_{n-1}]$

Hence the $n^{th}$ term of the A.P. will be $(6n-1)$.