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Anybody help me to solve this question. ​

Anybody help me to solve this question. ​-example-1

1 Answer

5 votes

Answer:


(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

User Waqas Ali
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