Find two numbers such that their mean proportional is 18 and the third proportional to them is 144.
$$\frac{x}{18}=\frac{18}{y}\ .$$
or, $$xy=18^2.$$...................(1)
Again,Â
$$\frac{x}{y}=\frac{y}{144}\ .$$
or, $$x=\frac{y^2}{144}\ .$$
So, from eq (1) :
$$\frac{18^2}{y}=\frac{y^2}{144}\ .$$
or, $$y^3=18^2\times144\ .$$
or, $$y=36.$$
So, $$x=\frac{18^2}{36}=9\ .$$
B is correct choice.
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