A) \[{{n}^{2}}:1\]
B) \[1:{{n}^{2}}\]
C) \[{{n}^{2/3}}:1\]
D) \[1:\,{{n}^{2/3}}\]
Correct Answer: C
Solution :
If A and B are mean and third proportional, then \[\frac{x}{A}=\frac{A}{y}\] or \[A=\sqrt{(xy)}\] and \[\frac{x}{y}=\frac{y}{B}\] or \[B=\frac{{{y}^{2}}}{x}\] Now, \[\sqrt{(xy)}=n\times \frac{{{y}^{2}}}{x}\] \[\Rightarrow \] \[{{\left( \frac{x}{y} \right)}^{3/2}}=\frac{n}{1}\Rightarrow \,\,\left( \frac{x}{y} \right)={{n}^{2/3}}:1\]
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