If the sum of n terms of an AP is

Solution:

Given:

$S_{n}=2 n^{2}+3 n$

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

= 5

$S_{2}=2(2)^{2}+3(2)$

= 14

$\therefore a_{1}+a_{2}=14$

$\Rightarrow 5+a_{2}=14$

$\Rightarrow a_{2}=9$

Common difference, $d=a_{2}-a_{1}$

= 9-5

= 4

nth term $=a+(n-1) d$

$=5+(n-1) 4$

$=4 n+1$