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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 :

- $ \frac{b}{2}$
- $ \frac{c}{2}$
- $ \frac{2}{b}$
- $ \frac{2}{a}$

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$Given$ : $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}$

$\textrm{Option C}$

$Let$ $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}$

$\textrm{Option C}$

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