Question: The sum of first $n$ terms of an A.P. is $3 n^{2}+4 n$. Find the 25 th term of this A.P.
Solution:
$S_{n}=3 n^{2}+4 n$
We know
$a_{n}=S_{n}-S_{n-1}$
$\therefore a_{n}=3 n^{2}+4 n-3(n-1)^{2}-4(n-1)$
$\Rightarrow a_{n}=6 n+1$
$a_{25}=6(25)+1=151$
