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If $x^{a}=y^{b}=z^{c}$ and $y^{2}=zx$ then the value of $\frac{1}{a} + \frac{1}{c}$ is  :

1. $\frac{b}{2}$
2. $\frac{c}{2}$
3. $\frac{2}{b}$
4. $\frac{2}{a}$

Given : $x^{a}=y^{b}=z^{c},y^2=zx$

Let  $x^{a} = y^{b} = z^{c}= k$

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

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

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

$\because y^{2}=zx$

$\therefore (k^{\frac{1}{b}})^2$ $=$ $(k^{\frac{1}{c}})*$ $(k^{\frac{1}{b}})$

$\therefore \frac{1}{a}+\frac{1}{b}=\frac{2}{b}$

Option $(C)$ is correct.
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