Aps

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

Gross Question ..

What I've done so far is:

$S_1 = \frac{n}{2} [2 + (n-1) ]$

$S_2 = \frac{n}{2}[4 + (n - 1)3 ]$

$S_3 = \frac{n}{2}[6 + (n - 1)5]$



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

What can I do from here??

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

Gross Question ..

What I've done so far is:

$S_1 = \frac{n}{2} [2 + (n-1) ]$

$S_2 = \frac{n}{2}[4 + (n - 1)3 ]$

$S_3 = \frac{n}{2}[6 + (n - 1)5]$



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

What can I do from here??

Have you tried plugging in values to your formulae? For your $S_1$ you get the sequence $1, 3, 6, \ldots$, which isn't correct, for example. We should have

$S_1 = 1 + (n-1)1 = n$

$S_2 = 2 + (n-1)3 = 3n-1$

$S_3 = 3 + (n-1)5 = 5n-2$

etc. etc. Summing all of these is a little less nasty!

I think you're getting mixed up with nth terms / summations.

The $S_1$ that the OP wrote is meant to represent the sum of the first sequence which is 1,2,3,4,...

This sequence has nth term $n$ but the sum of the first n terms of it is $\frac{n}{2} [2 + (n-1) ]$