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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}$
in Quantitative Aptitude recategorized by
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Given : $x^{a}=y^{b}=z^{c},y^2=zx$

Let  $x^{a} = y^{b} = z^{c}= k$

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

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

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

$\because y^{2}=zx$

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

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

Option $(C)$ is correct.
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