If the sum of first p term of an A.P. is ap2 + bp,

Here, we are given,

$S_{p}=a p^{2}+b p$

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

Now, as we know,

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

So, we get,

$a_{p}=\left(a p^{2}+b p\right)-\left[a(p-1)^{2}+b(p-1)\right]$

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

$=a p^{2}+b p-\left(a p^{2}+a-2 a p+b p-b\right)$

$=a p^{2}+b p-a p^{2}-a+2 a p-b p+b$

$=2 a p-a+b$$\ldots(1)$

Also,

$a_{p}=a^{\prime}+(p-1) d$

$=a^{\prime}+p d-d$

$=p d+\left(a^{\prime}-d\right)$$\ldots(2)$

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

$d p=2 a p$

$d=2 a$

Therefore, the common difference is $d=2 a$