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.
$\therefore$ $b=ar,c=a{r}^2$
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.