Arithmetic Progression - show that question

[trollcast]

Homework Statement

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.

Homework Equations

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 canceled out to 0 as both the sides worked out to be the same?

 

[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.

 

[SammyS]

Homework Statement

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.

Homework Equations

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 canceled out to 0 as both the sides worked out to be the same?

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 $\displaystyle \ \frac{1}{b+c}<\frac{1}{c+a}< \frac{1}{a+b} \ .$

 

[trollcast]

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

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

 

[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.

 

[haruspex]

check if $\frac{1}{a+c}$ - $\frac{1}{b+c}$ = $\frac{1}{a+b}$ - $\frac{1}{a+c}$.

As suggested in post #2.