A) \[\frac{b}{2}\]
B) \[\frac{c}{2}\]
C) \[\frac{2}{b}\]
D) 2a
Correct Answer: C
Solution :
If \[{{x}^{a}}={{y}^{b}}={{z}^{c}}\] and \[{{y}^{2}}=zx\] Let \[{{x}^{a}}={{y}^{b}}={{z}^{c}}=k\] \[\Rightarrow \] \[x={{k}^{1/b}},\,\,y={{k}^{1/b}},\,\,z={{k}^{1/c}}\] \[\Rightarrow \] \[{{y}^{2}}=zx\] \[\therefore \] \[{{({{k}^{1/b}})}^{2}}=({{k}^{1/c}})\cdot ({{k}^{1/a}})\] \[\Rightarrow \] \[{{(k)}^{2/b}}={{(k)}^{1/c+1/a}}\] \[\therefore \] \[\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\]
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