Q: Given p A.P’s, each of which consists of n terms. If their first terms are 1, 2, 3,……, p and common differences are 1, 3, 5, ….., 2p – 1 respectively , then sum of the terms of all the progressions is

(A) $ \frac{1}{2}np(np+1) $

(B) $ \frac{1}{2} n(p+1) $

(C) np(n+1)

(D) none of these

Ans: (A).

Sol: The r^{th} A. P. has first term r and common difference 2r-1.

Hence , sum of its n terms

$\large = \frac{n}{2} [2r + (n-1)(2r-1)] $

The required sum $ \large = \Sigma_{r=1}^{p} \frac{n}{2} [2r + (n-1)(2r-1)] $

$\large = n \Sigma_{r=1}^{p} r + \frac{n(n-1)}{2} [2 \Sigma_{r=1}^{p} r – \Sigma_{r=1}^{p} 1 ] $

$\large = \frac{n p(p+1)}{2} + \frac{n (n-1)}{2}[p(p+1)-p] $

$\large = \frac{1}{2}np(np + 1) $