# In an AP, the sum of n terms is $n^2+1$. Find the 16th term.

$\begin{array}{1 1} 17 \\ 257 \\ n^2 \\ 31 \end{array}$

Explanation : In an AP
$T_{n}=S_{n}-S_{n-1}\qquad\;S_{n}=n^2+1$
$T_{n}=n^2+1-[(n-1)^2+1]$
$=n^2+1-(n^2-2n+1+1)$
$=2n-1$
$T_{16}=2*16-1=32-1=31.$