Find two numbers such that their mean proportion is 16 and third proportion is 1024.

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 $$y$$ = 16

=> $$xy = (16)^2 = 256$$ ---------------(i)

Third proportion = 1024

=> $$y^2 = x \times 1024$$ -------------(ii)

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

=> $$y^2 = \frac{256}{y} \times 1024$$

=> $$y^3 = (2)^8 \times (2)^{10} = (2)^{18}$$

=> $$y = (2)^{18 \div 3} = (2)^6 = 64$$

Substituting it in equation(i), => $$x = \frac{256}{64} = 4$$