If `(b-c)^2,(c-a)^2,(a-b)^2` are in A.P., then prove that `1/(b-c),1/(c-a),1/(a-b)` are also in A.P.

Explanation

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

`2/(c-a)=1/(b-c)+1/(a-b)`

`2/(c-a)=(b-c+a-b)/((b-c)(a-b)`

`2[(1-c)(a-b)]=-(a-c)^2`

`2[-b-ac+ab+bc]=-a^2-c^2+2ac`

`a^2+c^2-4ac=b^2-2ab+b^2-2bc`

`2[a^2+c^2]-4ac=b^2-2ab+a^2+b^2-2bc+c^2`

`2(a-c)^2=(b-a)^2+(b-c)^2`.