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

(a) their product is 3375

(b) the result of the product of first and second number added to the product of second and third number is 750.

Find the numbers.

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

Given,

Product of the numbers is 3375.

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

(b) Given,

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

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

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

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

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

= 45, 15, 5.

Let r = 3

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

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

= 5, 15, 45.

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