Find two numbers such that their mean proportion is 12 and third proportion is 768.

Three numbers a,b,c are in proportion iff $$b^2 = ac$$ where $$b$$ is the mean proportion and $$c$$ is the third proportion

Mean proportion of two numbers $$x$$ and $$y$$ = 12

=> $$xy = (12)^2 = 144$$ ---------------(i)

Third proportion = 768

=> $$y^2 = x \times 768$$ -------------(ii)

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

=> $$y^2 = \frac{144}{y} \times 768$$

=> $$y^3 = (12)^3 \times (4)^3 = (12 \times 4)^3$$

=> $$y = 12 \times 4 = 48$$

Substituting it in equation(i), => $$x = \frac{144}{48} = 3$$