The sum of three numbers which form a geometric progression (G.P) is $13$ and the sum of their squares is $91$. Find the numbers.
Let the three numbers in G.P be $\;$ $\dfrac{a}{r}, \; a, \; a \cdot r$
where $r$ is the common ratio of the G.P
Given: $\;$ $\dfrac{a}{r} + a + a \cdot r = 13$
i.e. $\;$ $a \times \left(\dfrac{1}{r} + 1 + r\right) = 13$ $\;\;\; \cdots \; (1a)$
i.e. $\;$ $\dfrac{a}{r} \times \left(1 + r + r^2\right) = 13$
i.e. $\;$ $\dfrac{a^2}{r^2} = \dfrac{169}{\left(1 + r + r^2\right)^2}$ $\;\;\; \cdots \; (1b)$
And: $\;$ $\left(\dfrac{a}{r}\right)^2 + a^2 + \left(a \cdot r\right)^2 = 91$
i.e. $\;$ $a^2 \times \left(\dfrac{1}{r^2} + 1 + r^2\right) = 91$
i.e. $\;$ $\dfrac{a^2}{r^2} \times \left(1 + r^2 + r^4\right) = 91$
i.e. $\;$ $\dfrac{169}{\left(1 + r + r^2\right)^2} \times \left(1 + r^2 + r^4\right) = 91$ $\;\;\;$ [by equation $(1b)$]
i.e. $\;$ $\left(\dfrac{169}{1 + r + r^2}\right) \times \left(\dfrac{1 + r^2 + r^4}{1 + r + r^2}\right) = 91$
i.e. $\;$ $\left(\dfrac{169}{1 + r + r^2}\right) \times \left[r^2 + \dfrac{1 - r^3}{1 + r + r^2}\right] = 91$
i.e. $\;$ $\left(\dfrac{169}{1 + r + r^2}\right) \times \left[r^2 + \dfrac{\left(1 - r\right) \left(1 + r + r^2\right)}{1 + r + r^2}\right] = 91$
i.e. $\;$ $\left(\dfrac{169}{1 + r + r^2}\right) \times \left(r^2 - r + 1\right) = 91$
i.e. $\;$ $169 r^2 - 169 r + 169 = 91 + 91 r + 91 r^2$
i.e. $\;$ $78 r^2 - 260 r + 78 = 0$
i.e. $\;$ $3 r^2 - 10 r + 3 = 0$
i.e. $\;$ $\left(r - 3\right) \left(3r - 1\right) = 0$
i.e. $\;$ $r = 3$ $\;$ or $\;$ $r = \dfrac{1}{3}$
Substituting $r = 3$ in equation $(1a)$ gives
$a \times \left(\dfrac{1}{3} + 1 + 3\right) = 13$
i.e. $\;$ $a \times \dfrac{13}{3} = 13$ $\implies$ $a = 3$
Substituting $r = \dfrac{1}{3}$ in equation $(1a)$ gives
$a \times \left(\dfrac{1}{\dfrac{1}{3}} + 1 + \dfrac{1}{3}\right) = 13$
i.e. $\;$ $a \times \left(3 + 1 + \dfrac{1}{3}\right) = 13$
i.e. $\;$ $a \times \dfrac{13}{3} = 13$ $\implies$ $a = 3$
$\therefore \;$ When $a = 3$, $\;$ $r = 3$, the numbers are:
$\dfrac{3}{3}, \; 3 , \; 3 \times 3$ $\;$ i.e. $\;$ $1, \; 3, \; 9$
When $a = 3$, $\;$ $r = \dfrac{1}{3}$, the numbers are:
$\dfrac{3}{\dfrac{1}{3}}, \; 3 , \; 3 \times \dfrac{1}{3}$ $\;$ i.e. $\;$ $9, \; 3, \; 1$
$\implies$ the required numbers are $\;$ $1, \; 3, \; 9$