Question

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 .

Answer 1

Answer: (A)

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

Answer 2

The r^th A. P. has first term r and common difference 2r – 1. Hence sum of its n terms = $\frac{n}{2}\left\lbrack 2r + (n - 1)(2r - 1) \right\rbrack$.

The required sum = $\sum_{r = 1}^{p}{\frac{n}{2}\left\lbrack 2r + (n - 1)(2r - 1) \right\rbrack}$

= $n\sum_{r = 1}^{p}{r + \frac{n(n - 1)}{2}\left\lbrack 2\sum_{r = 1}^{p}{r - \sum_{r = 1}^{p}1} \right\rbrack}$

= $\frac{np(p + 1)}{2} + \frac{n(n - 1)}{2}\left\lbrack p(p + 1) - p \right\rbrack$

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