A) AP
B) GP
C) HP
D) None of these
Correct Answer: A
Solution :
Given,\[{{a}^{1/x}}={{b}^{1/y}}={{c}^{1/z}}\]and\[a,b,c\]are in GP. Let common ratio of GP is r. \[\therefore \] \[b=ar,\text{ }c=a{{r}^{2}}\] Now, \[{{a}^{1/x}}={{b}^{1/y}}={{c}^{1/z}}\] \[\Rightarrow \] \[\frac{1}{x}\log a=\frac{1}{y}\log b=\frac{1}{z}\log c\] \[\therefore \] \[\frac{y}{x}=\frac{\log b}{\log a}=\frac{\log a{{r}^{2}}}{\log a}=1+\frac{\log r}{\log a}\] ?..(i) Similarly, \[\frac{z}{x}=\frac{\log c}{\log a}=\frac{\log a{{r}^{2}}}{\log a}=1+\frac{2\log r}{\log a}\] ?.(ii) From Eqs. (i) and (ii), \[\frac{z}{x}=1+2\left( \frac{y}{x}-1 \right)\] \[\Rightarrow \] \[\frac{z}{x}=1+\frac{2y}{x}-2\] \[\Rightarrow \] \[\frac{z}{x}=\frac{2y}{x}-1\] \[\Rightarrow \] \[z=2y-x\] \[\Rightarrow \] \[2y=x+z\] Hence,\[x,y,z\]are in AP.
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