# Arithmetic Progression - show that question

1. Dec 14, 2012

### trollcast

1. The problem statement, all variables and given/known data

Given that a2, b2 and c 2 are in arithmetic progression show that:

$$\frac{1}{b+c} , \frac{1}{c+a} , \frac{1}{a+b}$$

,are also in arthimetic progression.

2. Relevant equations
3. The attempt at a solution

So I assume by "in arithmetic progression" they mean those are 3 consecutive terms but we can't assume that a2 is the first term?

Then,

$$b^{2} = a^{2} + d \\ c^2 = b^2 + d \\ c^2 = a^2 + 2d \\ d = b^{2} - a^{2} \\ d = c^{2} - b^{2}$$

However I can't seem to get anything close to what the questions asking me as when I tried solving that set equations I kept on getting equations that just cancelled out to 0 as both the sides worked out to be the same?

2. Dec 14, 2012

### haruspex

You can assume a < b < c. It follows that b+c > a+b > a+c, so you know what order the second sequence will be in.
Try it from the other end. Look at differences between consecutive terms of the second sequence, and take the difference of the differences.

3. Dec 14, 2012

### SammyS

Staff Emeritus
I would mess around with $\displaystyle \ \frac{1}{b+c} , \frac{1}{c+a} , \frac{1}{a+b} \$ just to see if I could find some connection with the squares of a, b, and c .

Although, it doesn't give an ordering, I would assume a < b < c .

Then c+b > c+a > b+a, so that $\displaystyle \ \frac{1}{b+c}<\frac{1}{c+a}< \frac{1}{a+b} \ .$

4. Dec 15, 2012

### trollcast

Found a similar problem on google but don't understand why they added (ab+ac+bc) to it although it does work.

$$a^2,b^2,c^2 \\ a^2+(ab+ac+bc),b^2+(ab+ac+bc),c^2+(ab+ac+bc) \\ a^2+ab+ac+bc,b^2+ab+ac+bc,c^2+ab+ac+bc \\ (a+b)(a+c),(b+a)(b+c),(c+a)(c+b) \\ \frac{(a+b)(a+c)}{(a+b)(a+c)(b+c)},\frac{(b+a)(b+c)}{(a+b)(a+c)(b+c)},\frac{(c+a)(c+b)}{(a+b)(a+c)(b+c)} \\ ∴ \frac{1}{(b+c)},\frac{1}{(a+c)},\frac{1}{(a+b)}$$

5. Dec 15, 2012

### kastelian

Given the a2, b2, c2 makes arithmetic progression that means b2 - a2 = c2 - b2.

For $\frac{1}{b+c}$, $\frac{1}{a+c}$, $\frac{1}{a+b}$ to make arithmetic progression it shall be enough to check if $\frac{1}{a+c}$ - $\frac{1}{b+c}$ = $\frac{1}{a+b}$ - $\frac{1}{a+c}$. Thanks for the ordering already established by previous posts.

And that easily simplifies to (b-a)(a+b)(a+c) = (c-b)(a+c)(b+c). That in turn is b2 - a2 = c2 - b2, which was given.

6. Dec 15, 2012

### haruspex

As suggested in post #2.