A) \[{{a}^{2}},\ {{b}^{2}},\ {{c}^{2}}\] are in G.P.
B) \[{{a}^{2}}(b+c),\ {{c}^{2}}(a+b),\ {{b}^{2}}(a+c)\] are in G.P.
C) \[\frac{a}{b+c},\ \frac{b}{c+a},\ \frac{c}{a+b}\] are in G.P.
D) None of the above
Correct Answer: A
Solution :
\[a={{r}^{2}}\] are in G.P. \[\therefore \ \frac{b}{a}=\frac{c}{b}=r\]\[\Rightarrow \]\[\frac{{{b}^{2}}}{{{a}^{2}}}=\frac{{{c}^{2}}}{{{b}^{2}}}={{r}^{2}}\] \[\Rightarrow \]\[{{a}^{2}},\,\ {{b}^{2}},\ {{c}^{2}}\]are in G.P.
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