FIITJEE A.P Problem

Algebra Level 2

Given $p$ arithmetic progressions, each of which consisting of $n$ terms, if their first terms are $1,2,3,\ldots,p-1,p$ and common differences are $1,3,5,7,\ldots,2p-3,2p-1,$ respectively, what is the sum of all the terms of all the arithmetic progressions?

np(np+1)

None of these

\dfrac{np}{2}(np+1)

\dfrac{n}{2}(p+1)

1 solution

Venkata Karthik Bandaru
Apr 13, 2015

Let us denote the sums of individual progressions as $S_{1}, S_{2},......,S_{p}$ . Then, we have :-

$S_{1} = \dfrac{n}{2} [2+(n-1)]$

$S_{2} = \dfrac{n}{2} [4+(n-1)3]$

$S_{3} = \dfrac{n}{2} [6+(n-1)5]$

..............so on till the last sum

$S_{p} = \dfrac{n}{2} [(2p-1)n+1]$

Hence, the required sum $= \dfrac{n}{2} [(n+1)+(3n+1)+.... + ((2p-1)(n-1))]$ $= \dfrac{n}{2} [ (n+3n+5n+..... +(2p-1)n)+p ]$ $= \dfrac{n}{2} [ n(1+3+5+....+(2p-1)) + p ]$ $= \dfrac{n}{2} [n p^{2}+p] =\boxed{\dfrac{np}{2} (np+1)}$ .

Moderator note:

Nicely done. Note that the sum of all these arithmetic progression sum is equals to $1 +2+3+\ldots + (np-1) + (np) = \frac {np} 2 (np + 1)$ .

Great solution! Although, I am confused by the sum n/2[(n + 1) + (3n + 1) + ... + ((2p - 1)(n - 1))]. Why wouldn't the last term in the series be (2p - 1)n + p, as you had written in your equation of Sp?

Christian Brown - 4 years, 10 months ago

FIITJEE A.P Problem

1 solution

Moderator note:

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