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  4. If b+c,c+a,a+b are in HP, then a2,b2,c2 are in:

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Q. If $ b+c,c+a,a+b $ are in HP, then $ {{a}^{2}},{{b}^{2}},{{c}^{2}} $ are in:

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Solution:

Since, $ b+c,c+a,a+b $ are in HP
$ \Rightarrow $ $ \frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b} $ are in AP
$ \Rightarrow $ $ \frac{1}{c+a}=\frac{1}{2}\left( \frac{1}{b+c}+\frac{1}{a+b} \right) $
$ \Rightarrow $ $ \frac{2}{c+a}=\frac{a+b+b+c}{(a+b)(b+c)} $
$ \Rightarrow $ $ (a+2\,b+c)(a+c)=2(a+b)(b+c) $
$ \Rightarrow $ $ {{a}^{2}}+ac+2\,ab+2\,bc+ac+{{c}^{2}} $
$ =2(ab+ac+{{b}^{2}}+bc) $
$ \Rightarrow $ $ {{a}^{2}}+{{c}^{2}}=2{{b}^{2}} $
$ \Rightarrow $ $ {{a}^{2}},{{b}^{2}},{{c}^{2}} $ are in AP