A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: C
Solution :
Given that \[{{b}^{2}},\ {{a}^{2}},\ {{c}^{2}}\] are in A.P. Therefore \[{{a}^{2}}-{{b}^{2}}={{c}^{2}}-{{a}^{2}}\] \[\Rightarrow \]\[(a-b)(a+b)=(c-a)(c+a)\] \[\Rightarrow \] \[\frac{a-b}{c+a}=\frac{c-a}{a+b}\]\[\Rightarrow \]\[\frac{b-a+c-c}{(c+a)(b+c)}=\frac{a+b-b-c}{(b+c)(a+b)}\] \[\Rightarrow \] \[\frac{1}{b+c}-\frac{1}{a+b}=\frac{1}{c+a}-\frac{1}{b+c}\] \[\Rightarrow \] \[\frac{1}{a+b},\ \frac{1}{b+c},\ \frac{1}{c+a}\] are in A.P. Hence \[(a+b),\ (b+c),\ (c+a)\] are in H.P.
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