Question

if (b-c)^2,(c-a)^2,(a-b)^2 are in AP, prove that 1/(b-c),1/(c-a),1/(a-b) are in AP

Answer 1

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Answer 2

Answer with 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