If $a^{x} = b^{y} = c^{z}$ and $a, b, c$ are G.P., then $x, y, z$ are in

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A.P

A.G.P

H.P

G.P

answer is C.

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Detailed Solution

$a^{x} = b^{y} = c^{z}$

$a, b, c$ are G.P.,

$b^{2} = a c$

$Let a^{x} = b^{y} = c^{z} = k$

$a^{x} = k \Rightarrow a = k^{\frac{1}{x}}$

$b^{y} = k \Rightarrow b = k^{\frac{1}{y}}$

$c^{x} = k \Rightarrow c = k^{\frac{1}{z}}$

${(k^{\frac{1}{y}})}^{2} = k^{\frac{1}{x}} . k^{\frac{1}{z}}$

$\Rightarrow k^{\frac{2}{y}} = k^{\frac{1}{x} + \frac{1}{z}}$

$\Rightarrow \frac{2}{y} = \frac{1}{x} - \frac{1}{z}$

$\Rightarrow \frac{1}{y} + \frac{1}{y} = \frac{1}{x} - \frac{1}{z}$

$\Rightarrow \frac{1}{y} - \frac{1}{x} = \frac{1}{z} - \frac{1}{y}$

$H_{3} - H_{2} = H_{4} - H_{3}$

$x, y, z$ are in H.P

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