Question

If $a^x=b^y=c^z=d^u$ and $a,b,c,d$ are in $GP$, then $x,y,z,u$ are in

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

Answer 1

Answer: (C)

We have,
$a^x=b^y=c^z=d^u$
Let $a^x=b^y=c^z=d^u=k$
$\Rightarrow a=k^{1/x},b=k^{1/y},c=k^{1/z},d=k^{1/u}$ ...(i)
Since, $a,b,c,d$ are in GP.
$\therefore \frac{b}{a}=\frac{c}{b}=\frac{d}{c}$
$\Rightarrow \frac{k^{1/y}}{k^{1/x}}=\frac{k^{1/z}}{k^{1/y}}=\frac{k^{1/u}}{k^{1/z}}$
$\{\text{using Eq}.(i)\}$
$\Rightarrow k^{\frac{1}{y}-\frac{1}{x}}=k^{\frac{1}{z}-\frac{1}{y}}=k^{\frac{1}{u}-\frac{1}{z}}$
$\Rightarrow \frac{1}{y}-\frac{1}{x}=\frac{1}{z}-\frac{1}{y}=\frac{1}{u}-\frac{1}{z}$
$\therefore \frac{1}{x},\frac{1}{y},\frac{1}{z},\frac{1}{u}$ are in AP.
$\Rightarrow x,y,z,u$ are in HP.