The sum of the first n terms of an AP is (3n2 + 6n).

Let Sn denotes the sum of first n terms of the AP.

$\therefore S_{n}=3 n^{2}+6 n$

$\Rightarrow S_{n-1}=3(n-1)^{2}+6(n-1)$

$=3\left(n^{2}-2 n+1\right)+6(n-1)$

$=3 n^{2}-3$

$\therefore n^{\text {th }}$ term of the AP, $a_{n}$

$=S_{n}-S_{n-1}$

$=\left(3 n^{2}+6 n\right)-\left(3 n^{2}-3\right)$

$=6 n+3$

Putting n = 15, we get

$a_{15}=6 \times 15+3=90+3=93$

Hence, the nth term is (6n + 3) and 15th term is 93.