If the sum of n terms of an A.P. is Sn = 3n2 + 5n.

Solution:

Here, we are given,

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

Let us take the first term as a and the common difference as d.

Now, as we know,

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

So, we get,

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

$=3 n^{2}+5 n-\left[3\left(n^{2}+1-2 n\right)+5 n-5\right]$ $\left[\right.$ Using $\left.(a-b)^{2}=a^{2}+b^{2}-a b\right]$

$=3 n^{2}+5 n-\left(3 n^{2}+3-6 n+5 n-5\right)$

$=3 n^{2}+5 n-3 n^{2}-3+6 n-5 n+5$

 

$=6 n+2$$\ldots(1)$

Also,

$a_{n}=a+(n-1) d$

$=a+n d-d$

 

$=n d+(a-d)$ ............(2)

On comparing the terms containing n in (1) and (2), we get,

$d n=6 n$

$d=6$

Therefore, the common difference is $d=6$.