A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: A
Solution :
Since \[{{a}^{2}},\ {{b}^{2}},\ {{c}^{2}}\] be in A.P. Then \[{{b}^{2}}-{{a}^{2}}={{c}^{2}}-{{b}^{2}}\] \[\Rightarrow \] \[(b-a)(b+a)=(c-b)(c+b)\] Þ \[\frac{b-a}{c+b}=\frac{c-b}{b+a}\] Þ \[\frac{(b-a)(a+b+c)}{(c+a)(b+c)}=\frac{(c-b)(a+b+c)}{(a+b)(c+a)}\] \[\Rightarrow \] \[\frac{{{b}^{2}}+bc-ac-{{a}^{2}}}{(c+a)(b+c)}=\frac{{{c}^{2}}+ac-ab-{{b}^{2}}}{(a+b)(c+a)}\] \[\Rightarrow \] \[\frac{b}{c+a}-\frac{a}{b+c}=\frac{c}{a+b}-\frac{b}{c+a}\] Hence \[\frac{a}{b+c},\ \frac{b}{c+a},\ \frac{c}{a+b}\] be in A.P.
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