There are three positive numbers in Geometric Progression (G.P.) such that :

(a) Let the three positive numbers be $\frac{a}{r}, a, ar$.

Given,

Product of the numbers is 3375.

$$\begin{aligned}\Rightarrow \frac{a}{r} \times a \times ar = 3375 \\[1em] \Rightarrow a^3 = 3375 \\[1em] \Rightarrow a^3 = (15)^3 \\[1em] \Rightarrow a = 15.\end{aligned}$$

(b) Given,

Result of the product of first and second number added to the product of second and third number is 750.

$$\begin{aligned}\therefore \frac{a}{r} \times a + a \times ar = 750 \\[1em] \Rightarrow \frac{a^2}{r} + a^2r = 750 \\[1em] \Rightarrow a^2\Big(\frac{1}{r} + r\Big) = 750 \\[1em] \Rightarrow 15^2\Big(\frac{1}{r} + r\Big) = 750 \\[1em] \Rightarrow \Big(\frac{1 + r^2}{r}\Big) = \frac{750}{15^2} \\[1em] \Rightarrow \Big(\frac{1 + r^2}{r}\Big) = \frac{10}{3} \\[1em] \Rightarrow 3(1 + r^2) = 10r \\[1em] \Rightarrow 3 + 3r^2 = 10r \\[1em] \Rightarrow 3r^2 - 10r + 3 = 0 \\[1em] \Rightarrow 3r^2 - 9r - r + 3 = 0 \\[1em] \Rightarrow 3r(r - 3) - 1(r - 3) = 0 \\[1em] \Rightarrow (3r - 1)(r - 3) = 0 \\[1em] \Rightarrow 3r - 1 = 0 \text{ or } r - 3 = 0 \\[1em] \Rightarrow 3r = 1 \text{ or } r = 3 \\[1em] \Rightarrow r = \frac{1}{3} \text{ or } r = 3.\end{aligned}$$

Let r = $\frac{1}{3}$

Numbers : $\frac{a}{r}, a, ar$

= $\frac{15}{\frac{1}{3}}, 15, 15 \times \frac{1}{3}$

= 45, 15, 5.

Let r = 3

Numbers : $\frac{a}{r}, a, ar$

= $\frac{15}{3}, 15, 15 \times 3$

= 5, 15, 45.

Hence, numbers are 5, 15 and 45 or 45, 15 and 5.