Very-Short and Short-Answer Questions

Solution:

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

$\therefore S_{n}=3 n^{2}+5 n$

$\Rightarrow S_{n-1}=3(n-1)^{2}+5(n-1)$

$=3\left(n^{2}-2 n+1\right)+5(n-1)$

$=3 n^{2}-n-2$

Now,

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

$=\left(3 n^{2}+5 n\right)-\left(3 n^{2}-n-2\right)$

$=6 n+2$

Let d be the common difference of the AP.

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

$=(6 n+2)-[6(n-1)+2]$

$=6 n+2-6(n-1)-2$

$=6$

Hence, the common difference of the AP is 6.