Problem about arithmetic progression

[Augustus58]

Hi,

can't solve following prob:
Let a, b and c be real numbers.
Given that a^2, b^2 and c^2 are in arithmetic progression show that 1 / (b + c), 1 / (c + a) and 1 / (a + b) are also in arithmetic progression.

From assumptions: b^2 = a^2 + nk and c^2 = b^2 + mk where k is some real number and n and m are whole numbers.

Then f.e. b^2 - a^2 = nk and 1 / (a + b) = (a - b) / (a^2 - b^2) =(b - a) / (nk).
Further 1 / (c + a) = (c - a) / ((k(m + n)) and 1 / (b + c) = (c - b) / (mk).

I tried to get something out from 1 / (b + c) - 1 / (c + a) and 1 / (c + a) - 1 / (a + b) related to definition of AP (f.e. pr and qr where p and q are whole numbers and r is real number).

Not solved yet :) Seems easy.. x)

 

[jedishrfu]

Welcome to PF!

This looks like a homework problem. PF uses a homework template where you write down the problem, state relevant formulas and then show some work. We can't help you much without some work shown.

To start things off:

Have you established that it is true?

Do you have a simple progression that illustrates the theorem ie values for a,b,c and m,n and k?

 

[willem2]

I wonder if it was meant that the numbers are meant to be consecutive members of an arithmetic progression, so m = b.

Work out the differences between 1 / (b + c) and 1 / (c + a) and also 1 / (c + a) and 1/(a+b). Make sure the fractions have the same divisors.

 

[Augustus58]

Hi,

prob solved. Thank You for your messages. It's definitely good to assume m = n :) One solution from another forum: http://mathforum.org/kb/message.jspa?messageID=9593624