Find two numbers such that the mean proportional between them is 14 and the third proportional to them is 112.

Let two numbers be x and y.

Given, 14 is mean proportional to x and y,

$\therefore \dfrac{x}{14} = \dfrac{14}{y} \\[1em] \Rightarrow xy = 196 \space …..(i)$

Given, 112 is third proportional to x and y,

$\therefore \dfrac{x}{y} = \dfrac{y}{112} \\[1em] \Rightarrow y^2 = 112x \\[1em] \Rightarrow x = \dfrac{y^2}{112} \space …..(ii)$

Substituting value of x from (ii) in (i) we get,

$\Rightarrow \dfrac{y^2}{112}.y = 196 \\[1em] \Rightarrow y^3 = 196 \times 112 \\[1em] \Rightarrow y^3 = 21952 \\[1em] \Rightarrow y = \sqrt[3]{21952} \\[1em] \Rightarrow y = 28.$

$x = \dfrac{28^2}{112} = \dfrac{784}{112}$ = 7.

Hence, numbers are 7 and 28.