Question

Very-Short and Short-Answer Questions

If the sum of first p terms of an AP is (ap² + bp), find its common difference. [CBSE 2010]

Answer 1

Let S_p denotes the sum of first p terms of the AP.
$$\begin{gathered} \therefore S_p=a{p}^2+bp \\ \Rightarrow S_{p-1}=a{\left(p-1\right)}^2+b\left(p-1\right) \\ =a\left(p^2-2p+1\right)+b\left(p-1\right) \\ =a{p}^2-\left(2a-b\right)p+\left(a-b\right) \end{gathered}$$
Now,
p^th term of the AP, a_p = S_p − S_{p − 1
$$\begin{gathered} =\left(a{p}^2+bp\right)-\left[a{p}^2-\left(2a-b\right)p+\left(a-b\right)\right] \\ =a{p}^2+bp-a{p}^2+\left(2a-b\right)p-\left(a-b\right) \\ =2ap-\left(a-b\right) \end{gathered}$$}
Let d be the common difference of the AP.
$$\begin{gathered} \therefore d=a_p-a_{p-1} \\ =\left[2ap-\left(a-b\right)\right]-\left[2a\left(p-1\right)-\left(a-b\right)\right] \\ =2ap-\left(a-b\right)-2a\left(p-1\right)+\left(a-b\right) \\ =2a \end{gathered}$$

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