Question

If $a^{1/x}=b^{1/y}=c^{1/z}$ and $a,b,c$are in geometric progression, then $x,y,z$ are in

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

Answer 1

The correct option is A

AP

Find the relation between $x$, $y$and $z$:

Given that $a^{1/x}=b^{1/y}=c^{1/z}$ and $a,b,c$are in geometric progression.

Let $a^{1/x}=b^{1/y}=c^{1/z}=p$

So,$a=p^x,b=p^y,c=p^z$

Since $a,b,c$are in geometric progression.

$\Rightarrow b^2=ac$

Putting the value of $a,b,c$ . We get

$$\begin{gathered} {\left(p^y\right)}^2=\left(p^x\right)\left(p^z\right) \\ \Rightarrow p^{2y}=p^{(x+z)} \end{gathered}$$

Equating powers on both sides. we get

$2y=x+z$

So $x$, $y$ and $z$ are in AP.

Hence, the correct option is (A).