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

 

[CellCoree]

I would first try to understand the problem and the given information. In this case, we are given three real numbers a, b, and c, and we are told that their squares are in arithmetic progression. This means that there is a common difference between each of their squares. We are then asked to show that the reciprocals of the sums of these numbers (1/(b+c), 1/(c+a), and 1/(a+b)) are also in arithmetic progression.

To solve this problem, we can start by using the given information to find relationships between the numbers. For example, we know that b^2 = a^2 + nk and c^2 = b^2 + mk, where k is a real number and n and m are whole numbers. We can then substitute these expressions into the reciprocals of the sums: 1/(b+c) = 1/((a+b)+c) = 1/(a+b) * 1/(1+c/(a+b)) = 1/(a+b) * 1/(1+(c^2-a^2)/(a^2+b^2)) = 1/(a+b) * 1/(1+(mk)/(nk)) = 1/(a+b) * 1/(1+m/n) = 1/(a+b) * (n/(n+m)). Similarly, we can find expressions for 1/(c+a) and 1/(a+b).

Next, we can use the definition of arithmetic progression (AP) to show that these expressions are also in AP. The definition of AP states that the difference between any two consecutive terms is constant. Therefore, if we take the difference between 1/(b+c) and 1/(c+a), we get 1/(b+c) - 1/(c+a) = 1/(a+b) * (n/(n+m)) - 1/(a+b) * (m/(n+m)) = 1/(a+b) * (n-m)/(n+m). This is a constant value, since n and m are whole numbers. Similarly, the difference between 1/(c+a) and 1/(a+b) is also a constant value.

Therefore, we have shown that the reciprocals of the sums of a, b, and c are in arithmetic progression. This is just one possible approach to solving this problem, and there may be