6 is the mean proportion between two numbers x and y and 48 is the third proportional of x and y. Find the numbers.

Let two numbers be x and y.

Given,

6 is mean proportion between x and y,

$\therefore \dfrac{x}{6} = \dfrac{6}{y} \\[1em] \Rightarrow xy = 36 \space …..(1)$

Given,

48 is third proportional to x and y,

$\therefore \dfrac{x}{y} = \dfrac{y}{48} \\[1em] \Rightarrow y^2 = 48x \\[1em] \Rightarrow x = \dfrac{y^2}{48} \space …..(2)$

Substituting value of x from equation (2) in (1) we get,

$\Rightarrow \dfrac{y^2}{48}.y = 36 \\[1em] \Rightarrow y^3 = 36 \times 48 \\[1em] \Rightarrow y^3 = 1728 \\[1em] \Rightarrow y = \sqrt[3]{1728} \\[1em] \Rightarrow y = 12.$

Substituting value of y in equation (2), we get :

$x = \dfrac{12^2}{48} = \dfrac{144}{48}$ = 3.

Hence, numbers are 3 and 12.