Question

Anybody help me to solve this question. ​

Answer 1

Answer:

$\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are\ in\ AP$

Step-by-step explanation:

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

To prove: $\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}$ 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

$\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}2\dfrac{1}{(c-a)} =\dfrac{1}{(b-c)}+\dfrac{1}{(a-b)}\\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-b+b-c}{(b-c)(a-b)} \\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-c}{(b-c)(a-b)} \\\\\Rightarrow2(b-c)(a-b) = (c-a)(a-c) \\\\\Rightarrow 2(ab-b^2-ac+bc)= -(a-c)^2\\\\\Rightarrow 2ab- 2b^2-2ac+2bc = -a^2-c^2+2ac\\\\\Rightarrow a^2-2b^2+c^2=4ac-2ab-2bc$

Which is the result of AP

.

Hence proved