Prove that $\frac{1}{b-c},\frac{1}{c-a},\frac{1}{a-b}$ are in arithmetic progression under given condition.

Question

Suppose $(b-c)^2,(c-a)^2,(a-b)^2$ are in arithmetic progression.

Then show that $\frac{1}{b-c},\frac{1}{c-a},\frac{1}{a-b}$ are also in an arithmetic progression.

Please help.

Answer 1

Since it is give that $(b-c)^2,(c-a)^2,(a-b)^2$ are in AP. So, you get that :

$$2(c-a)^2=(b-c)^2+(a-b)^2$$

$$\implies2c^2+2a^2-4ac=2b^2+c^2+a^2-2bc-2ab+2ac$$

$$\implies c^2+a^2-2ac=2b^2-2bc-2ab+2ac $$

$$\implies (c-a)^2=2(b^2-bc-ab+ac)$$

$$\implies (c-a)^2=-2(ab+bc-b^2-ac)$$

$$\implies (a-c)(c-a)=2(ab+bc-b^2-ac)$$

$$\implies (a-c)(c-a)=2(a-b)(b-c)$$

$$\implies \frac{(a-c)}{(a-b)(b-c)}=\frac{2}{(c-a)}$$

$$\implies \frac{(a-b)+(b-c)}{(a-b)(b-c)}=\frac{2}{(c-a)}$$ $$\implies \frac{1}{(b-c)}+\frac{1}{(a-b)}=\frac{2}{(c-a)}$$

That is what we needed.