Question

The sum of first n terms of of an AP is given by Sn=2nsquare+3n.Find the sixteenth term

Answer 1

see solution in picture and hope it is correct

Answer 2

The sum of first n terms of an AP is given by
$S_n=2n^2+3n$.

we know, $T_n=S_n-S_{n-1}$
here, $S_n=2n^2+3n$
$S_{n-1}=2(n-1)^2+3(n-1)$

$T_n=2n^2+3n-[2(n-1)^2+3(n-1)]\\\\=2n^2-2(n-1)^2+3n-3(n-1)\\\\=2[n^2-n^2+2n-1]+3[n-n+1]\\\\=2(2n-1)+3\\\\=4n-2+3=4n+1$

so, 16th term ,$T_{16}=4\times16+1$

$T_{16}=64+1=65$