Find two numbers whose mean proportional is 16 and the third proportional is 128.

Let the two numbers be a and b.

Given, mean proportional between a and b is 16.

$\therefore \sqrt{ab} = 16 \qquad \text{[….Eq 1]}$

Given, third proportional is 128.

$\therefore \dfrac{a}{b} = \dfrac{b}{128} \\[0.5em] \Rightarrow a = \dfrac{b^2}{128} \qquad \text{[….Eq 2]}$

Putting this value of a in Eq 1,

$\Rightarrow \sqrt{\Big(\dfrac{b^2}{128}\Big)b} = 16 \\[0.5em] \Rightarrow \sqrt{\Big(\dfrac{b^3}{128}\Big)} = 16$

Squaring both sides,

$\Rightarrow \dfrac{b^3}{128} = 256 \\[1em] \Rightarrow b^3 = 256 \times 128 \\[1em] \Rightarrow b^3 = 32768 \\[1em] \Rightarrow b = \sqrt[3]{32768} \\[1em] \Rightarrow b = 32$

Putting value of b in Eq 2, $a = \dfrac{b^2}{128} \qquad \text{[….Eq 2]} \\[1em] \Rightarrow a = \dfrac{(32)^2}{128} \\[1em] \Rightarrow a = \dfrac{1024}{128} \\[1em] \Rightarrow a = 8$

Hence, the value of a = 8 and b = 32.