Find two numbers such that the mean proportion between them is 12 and the third proportional to them is 96.

Let the numbers be x and y.

Given, mean proportion = 12

∴ $\dfrac{x}{12} = \dfrac{12}{y} \Rightarrow xy = 144.$

Given, third proportion = 96

∴ $\dfrac{x}{y} = \dfrac{y}{96} \Rightarrow x = \dfrac{y^2}{96}$

Substituting value of x in xy = 144 we get,

$\Rightarrow \dfrac{y^2}{96}.y = 144 \\[1em] \Rightarrow \dfrac{y^3}{96} = 144 \\[1em] \Rightarrow y^3 = 144 \times 96 \\[1em] \Rightarrow y^3 = 13824 \\[1em] \Rightarrow y = 24.$

x = $\dfrac{y^2}{96} = \dfrac{24^2}{96} = 6.$

Hence, numbers are 6 and 24.