A) \[\frac{xz}{x+y}\]
B) \[\frac{xz}{2(x-z)}\]
C) \[\frac{xz}{2(z-x)}\]
D) \[\frac{2xz}{z+x}\]
Correct Answer: D
Solution :
\[{{a}^{x}}={{b}^{y}}={{c}^{z}}={{k}^{y}}\] (say) \[\Rightarrow \,\,\,a={{k}^{1/x}},\,\,b={{k}^{1/y}},\,\,c={{k}^{1/z}}\] \[\because \,\,\,\,\,{{b}^{2}}=ac\Rightarrow {{({{k}^{1/y}})}^{2}}={{k}^{1/x}}.{{k}^{1/z}}\] \[\Rightarrow \,\,\,\,\,{{k}^{2/y}}={{k}^{(1/x)+(1/z)}}\Rightarrow \frac{2}{y}=\frac{x+z}{xz}\] \[\therefore \,\,\,\,\,\,y=\]
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