A) AP
B) GP
C) HP
D) None of these
Correct Answer: C
Solution :
\[{{a}^{x}}={{b}^{y}}={{c}^{z}}={{d}^{u}}=k\] (suppose) \[\therefore \]\[a={{k}^{1/x}},b={{k}^{1/y}},c={{k}^{1/z}},d={{k}^{1/u}}\] Since a, b, c, d are in GP. \[\therefore \] \[{{b}^{2}}=ac\] \[\Rightarrow \] \[{{k}^{2/y}}={{k}^{1/x}}{{k}^{1/z}}\] \[\Rightarrow \] \[{{k}^{2/y}}={{k}^{1/x+1/z}}\] \[\Rightarrow \] \[\frac{2}{y}=\frac{1}{x}+\frac{1}{z}\] \[\therefore \]\[x,y,z\] are in HP.
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