Question

If $ {{a}^{1/x}}={{b}^{1/y}}={{c}^{1/z}} $ and $ a,b,c $ are in GP, then $ x,y,z $ will be in

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

Answer: (A)

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.