A) a, b, c are in AP
B) c, a, b are hi AP
C) \[{{a}^{2}},\text{ }{{b}^{2}},\text{ }{{c}^{2}}\]are in AP
D) a, b, care in GP
E) \[{{c}^{2}},\text{ }{{a}^{2}},\text{ }{{b}^{2}}\]are in AP
Correct Answer: C
Solution :
\[\because \]\[\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}\]are in AP . \[\therefore \]\[\frac{b}{c+a}-\frac{a}{b+c}=\frac{c}{a+b}-\frac{b}{c+a}\] \[\Rightarrow \]\[\frac{{{b}^{2}}+bc-ac-{{a}^{2}}}{(c+a)(b+c)}=\frac{{{c}^{2}}+ac-ab-{{b}^{2}}}{(a+b)(c+a)}\] \[\Rightarrow \]\[\{{{b}^{2}}-{{a}^{2}}-c(a-b)\}(a+b)\] \[=({{c}^{2}}-{{b}^{2}}-a(b-c)\}(b+c)\] \[\Rightarrow \]\[({{b}^{2}}-{{a}^{2}})(b+a+c)\] \[=({{c}^{2}}-{{b}^{2}})(a+b+c)\] \[\Rightarrow \] \[2{{b}^{2}}={{a}^{2}}+{{c}^{2}}\] \[\Rightarrow \]\[{{a}^{2}},\text{ }{{b}^{2}},\text{ }{{c}^{2}}\]are in AP.
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