Question

Anybody help me to solve this question. ​

Answer 1

`(1)/((b-c)),(1)/((c-a)) ,(1)/((a-b)) are\ in\ AP`

Explanation:

Given that
`(b-c)^2, (c-a)^2 , (a-b)^2` are in AP

To prove:
`(1)/((b-c)),(1)/((c-a)) ,(1)/((a-b))` are in AP

From given as we know if p , q, r are in AP then 2q= p+r.

`2(c-a)^2= (b-c)^2+(a-b)^2\\\\\Rightarrow 2(c^2+a^2-2ac)=b^2+c^2-2bc+a^2+b^2-2ab\\\\\Rightarrow 2c^2+2a^2-4ac= 2b^2+c^2+a^2 -2bc-2ab\\\\\Rightarrow a^2+c^2-2b^2-4ac= -2bc-2ab\\\\\Rightarrow a^2-2b^2+c^2= 4ac-2bc-2ab`

Now

`(1)/((b-c)),(1)/((c-a)) ,(1)/((a-b))2(1)/((c-a)) =(1)/((b-c))+(1)/((a-b))\\\\\Rightarrow (2)/((c-a))= (a-b+b-c)/((b-c)(a-b)) \\\\\Rightarrow (2)/((c-a))= (a-c)/((b-c)(a-b)) \\\\\Rightarrow2(b-c)(a-b) = (c-a)(a-c) \\\\\Rightarrow 2(ab-b^2-ac+bc)= -(a-c)^2\\\\\Rightarrow 2ab- 2b^2-2ac+2bc = -a^2-c^2+2ac\\\\\Rightarrow a^2-2b^2+c^2=4ac-2ab-2bc`

Which is the result of AP

.

Hence proved