A) AP
B) GP
C) HP
D) None of these
Correct Answer: C
Solution :
We have, \[{{a}^{x}}={{b}^{y}}={{c}^{z}}={{d}^{u}}\] Let \[{{a}^{x}}={{b}^{y}}={{c}^{z}}={{d}^{u}}=k\] \[\Rightarrow \]\[a={{k}^{1/x}},b={{k}^{1/y}},c={{k}^{1/z}},d={{k}^{1/u}}\] Since, \[a,\,b,\,c,\,d\]are in GP. \[\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\] \[\Rightarrow \] \[\frac{{{k}^{1/y}}}{{{k}^{1/x}}}=\frac{{{k}^{1/z}}}{{{k}^{1/y}}}=\frac{{{k}^{1/u}}}{{{k}^{1/z}}}\] {using Eq. (i)} \[\Rightarrow \] \[{{k}^{\frac{1}{y}-\frac{1}{x}}}={{k}^{\frac{1}{z}-\frac{1}{y}}}={{k}^{\frac{1}{u}-\frac{1}{z}}}\] \[\Rightarrow \] \[\frac{1}{y}-\frac{1}{x}=\frac{1}{z}-\frac{1}{y}=\frac{1}{y}-\frac{1}{z}\] \[\therefore \] \[\frac{1}{x},\frac{1}{y},\frac{1}{z},\frac{1}{u}\]are in AP. \[\Rightarrow \]\[x,y,z,u\]are in HP.
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