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)

HP

Answer 2

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.