Very-Short and Short-Answer Questions

Solution:

Let Sp denotes the sum of first p terms of the AP.

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

$\Rightarrow S_{p-1}=a(p-1)^{2}+b(p-1)$

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

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

Now,

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

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

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

$=2 a p-(a-b)$

Let d be the common difference of the AP.

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

$=[2 a p-(a-b)]-[2 a(p-1)-(a-b)]$

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

$=2 a$

Hence, the common difference of the AP is 2a.