Question

The sum of first $n$ terms of an A.P is given by $S_n=2n^2+3n$. Find the sixteenth term of the A.P

Answer 1

$Sn=\frac{n}{2}(2a+(n-1\left)d\right)$

$=\frac{n}{2}(2a-d)+n^2\frac{d}{2}=2n^2+3n$

$\therefore 2=\frac{d}{2}\Rightarrow =4$ and $\frac{2a-d}{2}=3$

$\therefore 2a=6+d=10$

or, $a=5$.

$\therefore a_{16}=a+(1+-1)d$

$=5+15\times 4=65.$