The sum of three numbers in a geometric progression is $\dfrac{39}{10}$ and their product is $1$. Find the numbers.
Let the three numbers in GP be $\;$ $\dfrac{a}{r}, \; a, \; ar$
where $\;$ $a = $ first term of GP and $\;$ $r = $ common ratio of GP
As per question,
$\dfrac{a}{r} + a + ar = \dfrac{39}{10}$ $\;\;\; \cdots \; (1)$
and $\;$ $\dfrac{a}{r} \times a \times ar = 1$ $\implies$ $a^3 = 1$ $\implies$ $a = 1$
Substituting the value of $\;$ $a$ $\;$ in equation $(1)$ we have,
$\dfrac{1}{r} + 1 + r = \dfrac{39}{10}$
i.e. $\;$ $\dfrac{1 + r^2}{r} = \dfrac{39}{10} - 1 = \dfrac{29}{10}$
i.e. $\;$ $10 + 10 r^2 = 29 r$
i.e. $\;$ $10 r^2 - 29 r + 10 = 0$
i.e. $\;$ $10 r^2 - 25 r - 4 r + 10 = 0$
i.e. $\;$ $5r \left(2r - 5\right) - 2 \left(2r - 5\right) = 0$
i.e. $\;$ $\left(5r - 2\right) \left(2r - 5\right) = 0$
i.e. $\;$ $5r - 2 = 0$ $\;$ or $\;$ $2r - 5 = 0$
i.e. $\;$ $r = \dfrac{2}{5}$ $\;$ or $\;$ $r = \dfrac{5}{2}$
$\therefore \;$ The three numbers in GP are:
$\dfrac{1}{2/5}, \; 1, \; 1 \times \dfrac{2}{5}$ $\;$ or $\;$ $\dfrac{1}{5/2}, \; 1, \; 1 \times \dfrac{5}{2}$
i.e. $\;$ $\dfrac{5}{2}, \; 1, \; \dfrac{2}{5}$ $\;$ or $\;$ $\dfrac{2}{5}, \; 1, \; \dfrac{5}{2}$