A) \[{{a}^{2}},\text{ }{{b}^{2}},\text{ }{{c}^{2}}\] are in GP.
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 these
Correct Answer: A
Solution :
[a] \[\because \] a, b, c are in GP. \[\therefore \,\,\frac{b}{a}=\frac{c}{b}=r\Rightarrow \frac{{{b}^{2}}}{{{a}^{2}}}=\frac{{{c}^{2}}}{{{b}^{2}}}={{r}^{2}}\] \[\Rightarrow \,\,{{a}^{2}},{{b}^{2,}}{{c}^{2}}\] are in GP.
You need to login to perform this action.
You will be redirected in
3 sec
