Find the mean proportion of :

(i) 5 and 80

(ii) $\dfrac{1}{12}$ and $\dfrac{1}{75}$

(iii) 8.1 and 2.5

(iv) (a - b) and (a³ - a²b), a > b.

(i) Let the mean proportion be x.

$\therefore \dfrac{5}{x} = \dfrac{x}{80} \\[0.5em] \Rightarrow x^2 = 5 \times 80 \\[0.5em] \Rightarrow x^2 = 400 \\[0.5em] \Rightarrow x = \sqrt{400} \\[0.5em] \Rightarrow x = 20.$

Hence, the mean proportion is 20.

(ii) Let the mean proportion be x.

$\therefore \dfrac{\dfrac{1}{12}}{x} = \dfrac{x}{\dfrac{1}{75}} \\[0.5em] \Rightarrow x^2 = \dfrac{1}{12} \times \dfrac{1}{75} \\[0.5em] \Rightarrow x^2 = \dfrac{1}{900} \\[0.5em] \Rightarrow x = \sqrt{\dfrac{1}{900}} \\[0.5em] \Rightarrow x = \dfrac{1}{30}$

Hence, the mean proportion is $\dfrac{1}{30}$.

(iii) Let the mean proportion be x.

$\therefore \dfrac{8.1}{x} = \dfrac{x}{2.5} \\[0.5em] \Rightarrow x^2 = 8.1 \times 2.5 \\[0.5em] \Rightarrow x^2 = 20.25 \\[0.5em] \Rightarrow x = \sqrt{20.25} \\[0.5em] \Rightarrow x = 4.5.$

Hence, the mean proportion is 4.5.

(iv) Let the mean proportion be x.

$\therefore \dfrac{(a - b)}{x} = \dfrac{x}{(a^3 - a^2b)} \\[0.5em] \Rightarrow x^2 = (a - b) \times (a^3 - a^2b) \\[0.5em] \Rightarrow x^2 = (a^4 - a^3b - a^3b + a^2b^2) \\[0.5em] \Rightarrow x^2 = (a^4 - 2a^3b + a^2b^2) \\[0.5em] \Rightarrow x^2 = a^2(a^2 - 2ab + b^2) \\[0.5em] \Rightarrow x^2 = a^2(a - b)^2 \\[0.5em] \Rightarrow x = \sqrt{(a^2(a - b)^2)} \\[0.5em] \Rightarrow x = a(a - b).$

Hence, the mean proportion is a(a - b).