The sum of three numbers in G.P. is $\dfrac{39}{10}$ and their product is 1. Find the numbers.

Let the numbers be $\dfrac{a}{r}, a, ar$.

Given,

Product = 1.

$\therefore \dfrac{a}{r} \times a \times ar = 1 \\[1em] \Rightarrow a^3 = 1 \\[1em] \Rightarrow a = 1.$

Sum = $\dfrac{39}{10}$

$\therefore \dfrac{a}{r} + a + ar = \dfrac{39}{10} \\[1em] \Rightarrow \dfrac{1}{r} + 1 + r = \dfrac{39}{10} \\[1em] \Rightarrow \dfrac{1 + r + r^2}{r} = \dfrac{39}{10} \\[1em] \Rightarrow 10(1 + r + r^2) = 39r \\[1em] \Rightarrow 10 + 10r + 10r^2 = 39r \\[1em] \Rightarrow 10r^2 - 29r + 10 = 0 \\[1em] \Rightarrow 10r^2 - 25r - 4r + 10 = 0 \\[1em] \Rightarrow 5r(2r - 5) - 2(2r - 5) = 0 \\[1em] \Rightarrow (5r - 2)(2r - 5) = 0 \\[1em] \Rightarrow 5r - 2 = 0 \text{ or } 2r - 5 = 0 \\[1em] \Rightarrow r = \dfrac{2}{5} \text{ or } r = \dfrac{5}{2}$

Let r = $\dfrac{2}{5}$

Numbers = $\dfrac{a}{r} = \dfrac{1}{\dfrac{2}{5}} = \dfrac{5}{2}$

a = 1

ar = $1 \times \dfrac{2}{5} =\dfrac{2}{5}$.

Let r = $\dfrac{5}{2}$

Numbers = $\dfrac{a}{r} = \dfrac{1}{\dfrac{5}{2}} = \dfrac{2}{5}$

a = 1

ar = $1 \times \dfrac{5}{2} =\dfrac{5}{2}$.

Hence, numbers = $\dfrac{5}{2}, 1, \dfrac{2}{5} \text{ or } \dfrac{2}{5}, 1, \dfrac{5}{2}$.