Choose the correct answer in each of the following questions:

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

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

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

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

$=3 n^{2}-3$

So,

$n^{\text {th }}$ term of the $\mathrm{AP}, a_{n}=S_{n}-S_{n-1}$

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

$=6 n+3$

Let d be the common difference of the AP.

$\therefore d=a_{n}-a_{n-1}$

$=(6 n+3)-[6(n-1)+3]$

$=6 n+3-6(n-1)-3$

$=6$

Thus, the common difference of the AP is 6.

Hence, the correct answer is option A.