Series, arithmatic progression.

[psyclone]

Homework Statement

Hi All,
I found this problem,

The sum of p, q, r terms of an Arithmetic Progression, are P, Q, R respectively: show that$$\frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0$$

Homework Equations

3. The Attempt at a Solution [/B]

My thoughts on how to start the problem is;

if
$$S_{n} = \frac{a}{2} (n + (n-1)d )$$

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

$$P = S_{p} = \frac{a}{2} (p + (p-1)d )$$

Therefore;

$$Q = S_{q} = \frac{a}{2} (q + (q-1)d )$$$$R = S_{r} = \frac{a}{2} (r + (r-1)d )$$

If I used the following series, to simplify a little, $$S_{n} = 1 + 2 + 3 ... + n,$$ then $$S_{n} = \frac{1}{2}n(n+1)$$

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

 

[mfb]

I guess you can just plug in those expressions for the sums and simplify.

 

[psyclone]

Solved it!

$$\frac{P}{p} = \frac{1}{2}(a + d(p-1))$$ $$(eq 1)$$

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

$$a = 2\frac{Q}{q} - d(q-1)$$ $$(eq 2)$$

Sub, eq 1 into eq 2

$$d = 2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} )$$

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

$$d = 2 \frac{Q}{(p-r)} ( \frac{Q}{q} - \frac{R}{r} )$$

$$2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} ) = 2 \frac{Q}{(p-r)} ( \frac{Q}{q} - \frac{R}{r} )$$

Simplify,

$$\frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0$$