The mean proportion between two numbers is 6 and their third proportion is 48. Find two numbers.

Let the numbers be a and b.

Given,

Mean proportion between the numbers is 6.

$\therefore \dfrac{a}{6} = \dfrac{6}{b} \\[1em] \Rightarrow ab = 36 \\[1em] \Rightarrow a = \dfrac{36}{b}\text{ ………(1)}$

Third proportion of the numbers is 48.

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

Substituting value of b in equation (1), we get :

$\Rightarrow a = \dfrac{36}{12}$ = 3.

Hence, the numbers are 3 and 12.