Series

[psyclones]

 

Hi All,
I found this problem,

The sum of p, q, r terms of an Arithmetic Progression, are P, Q, R respectively: show that


[latex] \frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0 [/latex]

 

 

My thoughts on how to start the problem is;

if
[latex] S_{n} = \frac{a}{2} (n + (n-1)d ) [/latex]


then the sum of say 'p' terms, would be


[latex] P = S_{p} = \frac{a}{2} (p + (p-1)d ) [/latex]


Therefore;

[latex] Q = S_{q} = \frac{a}{2} (q + (q-1)d ) [/latex]


[latex] R = S_{r} = \frac{a}{2} (r + (r-1)d ) [/latex]

If I used the following series, to simplify the above P, Q & R series; [latex] S_{n} = 1 + 2 + 3 ... + n, [/latex] then [latex] S_{n} = \frac{1}{2}n(n+1) [/latex]

But how to form the above equation, which contains all the terms (i.e, p, q, r, P, Q & R)?

Edited February 15, 2015 by psyclones

[imatfaal]

S_n = n/2(2a_0+(n-1)d)

 

I think you have your n and a_0 switched at first glance

yes - it works

You have an equation in P Q R and p q r - and you can set up P in terms of p a and d etc. If you correctly set up the sum equation - then sub in three times with expressions in a, d, p,r, and q. Simplify. It works

 

If you struggle I have done it and have a hand-written confirmation - it's a bit too lengthy to change into latex. So I cannot give you a line at a time - just paste the whole thing as a jpg.

[psyclones]

Thank-you for your post, I solved it!


[latex] \frac{P}{p} = \frac{1}{2}(a + d(p-1)) [/latex] [latex] (eq 1) [/latex]

Use sum of Q, to let a be the subject.

[latex] a = 2\frac{Q}{q} - d(q-1) [/latex] [latex] (eq 2) [/latex]

Sub, eq 1 into eq 2

[latex] d = 2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} ) [/latex]

Do the same for Sum Q & R, solve for d.

[latex] d = 2 \frac{Q}{(p-r)} ( \frac{Q}{q} - \frac{R}{r} ) [/latex]

[latex] 2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} ) = 2 \frac{Q}{(p-r)} ( \frac{Q}{q} - \frac{R}{r} ) [/latex]

Simplify,

[latex] \frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0 [/latex]

[imatfaal]