Find two numbers such that their mean proportion is 8 and third proportion is 64.

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 $$y$$ = 8

=> $$xy = (8)^2 = 64$$ ---------------(i)

Third proportion = 64

=> $$y^2 = x \times 64$$ -------------(ii)

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

=> $$y^2 = \frac{64}{y} \times 64$$

=> $$y^3 = (2)^6 \times (2)^6 = (2)^{6 + 6}$$

=> $$y = (2)^{\frac{12}{3}} = 2^4 = 16$$

Substituting it in equation(i), => $$x = \frac{64}{16} = 4$$

=> Ans - (D)