A) AP
B) HP
C) GP
D) none of these
Correct Answer: A
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 \[\Rightarrow \] \[\frac{1}{c+a}=\frac{1}{2}\left( \frac{1}{b+c}+\frac{1}{a+b} \right)\] \[\Rightarrow \] \[\frac{2}{c+a}=\frac{a+b+b+c}{(a+b)(b+c)}\] \[\Rightarrow \]\[(a+2b+c)(a+c)=2(a+b)(b+c)\] \[\Rightarrow \]\[{{a}^{2}}+ac+2ab+2bc+ac+{{c}^{2}}\] \[=2(ab+ac+{{b}^{2}}+bc)\] \[\Rightarrow \]\[{{a}^{2}}+{{c}^{2}}=2{{b}^{2}}\] \[\Rightarrow \]\[{{a}^{2}},{{b}^{2}},{{c}^{2}}\]are in AP
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