x and y are two numbers such that their mean proportion is 16 and third proportion is 128. What is the value of x and y?

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 = 128

=> $$y^2 = x \times 128$$ -------------(ii)

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

=> $$y^2 = \frac{256}{y} \times 128$$

=> $$y^3 = (2)^8 \times (2)^7 = (2)^{8 + 7}$$

=> $$y = (2)^{\frac{15}{3}} = 2^5 = 32$$

Substituting it in equation(i), => $$x = \frac{256}{32} = 8$$