Question

If the sum of n terms of an A.P is $2n^2+3n$, then write its nth term.

Answer 1

We have,

$S_n=2n^2+3n$

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

$=2(n^2+1-2n)+3n-3$

$=2n^2+2-4n+3n-3$

$=2n^2-n-1$

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

$\therefore T_n=S_n-S_{n-1}$

$=2n^2+3n-(2n^2-n-1)$

$=2n^2+3n-2n^2+n+1$

$=4n+1$

Hence, $T_n=4n+1$