Question

If 1/a+b;,1/b+c,1/c+a are in AP then prove that a^2,b^2,c^2 are also I. A.p

Answer 1

I hope this helps you.

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Answer 2

Answer:

Consider 1/a+b,1/b+c,1/c+a are in AP

So we know that in AP 2nd term-1st term

So we have $\frac{1}{b+c}-\frac{1}{a+b}=\frac{1}{c+a}-\frac{1}{b+c}$

By taking LCM for the denominators of the above expression and simplifying

We get a+b – (b+c)/(b+c)(a+b)=b+c-(c+a)/(c+a)(b+c)

By cancellation of the known terms

$=\frac{a-c}{a+b}=\frac{b-a}{c+a}$

By cross multiplication

=(a-c)(c+a)=(a+b)(b-a)

$=a c+a^{2}-c^{2}-c a=a b-a^{2}+b^{2}-b a$  [by multiplication of the terms]

By cancelling the terms we get

$=a^{2}-c^{2}=-a^{2}+b^{2}$

Thus the terms $\bold{a^{2}, b^{2} \text { and } c^{2}}$ are in AP

Hence proved