Find two numbers such that the mean proportional between them is 28 and the third proportional to them is 224.

Let the two numbers be x and y.

Given, 28 is the mean proportional between x and y.

$\therefore \dfrac{x}{28} = \dfrac{28}{y} \\[0.5em] \Rightarrow xy = 28 \times 28 \\[0.5em] \Rightarrow xy = 784 \\[0.5em] \Rightarrow x = \dfrac{784}{y} \qquad \text{[….Eq 1]}$

Given, 224 is the third proportional to numbers.

$\therefore \dfrac{x}{y} = \dfrac{y}{224} \\[0.5em] \Rightarrow y^2 = 224x \\[0.5em]$

Putting value of x from equation 1 above:

$\Rightarrow y^2 = 224\Big(\dfrac{784}{y}\Big) \\[0.5em] \Rightarrow y^3 = 175616 \\[0.5em] \Rightarrow y = \sqrt[3]{175616} \\[0.5em] \Rightarrow y = \sqrt[3]{56^3} \\[0.5em] \Rightarrow y = 56 \\[0.5em] \text{ and } x = \dfrac{784}{56} = 14.$

Hence, the two numbers are 14 and 56.