A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: A
Solution :
Let \[{{a}^{1/x}}={{b}^{1/y}}={{c}^{1/z}}=k\Rightarrow a={{k}^{x}},\,b={{k}^{y}},\ c={{k}^{z}}\] Now, \[a,\ b,\ c\]are in G.P. \[\Rightarrow \] \[{{b}^{2}}=ac\Rightarrow {{k}^{2y}}={{k}^{x}}.{{k}^{z}}={{k}^{x+z}}\Rightarrow 2y=x+z\] \[\Rightarrow \] \[x,\ y,\ z\] are in A.P.You need to login to perform this action.
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