Find three numbers which form a geometric progression (G.P) if their sum is $35$ and the sum of their squares is $525$.
Let the first term of the G.P be $= t_1 = a$
and the common ratio be $= r$
Let the three numbers in G.P be $\;$ $\dfrac{a}{r}, \; a, \; a \cdot r$
Given: $\;$ $\dfrac{a}{r} + a + a \cdot r = 35$
i.e. $\;$ $a \left(\dfrac{1}{r} + 1 + r\right) = 35$
$\implies$ $a = \dfrac{35 r}{1 + r + r^2}$ $\;\;\; \cdots \; (1)$
And: $\;$ $\left(\dfrac{a}{r}\right)^2 + a^2 + \left(a \cdot r\right)^2 = 525$
i.e. $\;$ $a^2 \times \left(\dfrac{1}{r^2} + 1 + r^2\right) = 525$ $\;\;\; \cdots \; (2)$
Substituting the value of $a$ from equation $(1)$ in equation $(2)$ gives
$\left(\dfrac{35 r}{1 + r + r^2}\right)^2 \times \left(\dfrac{1}{r^2} + 1 + r^2\right) = 525$
i.e. $\;$ $\dfrac{1 + r^2 + r^4}{\left(1 + r + r^2\right)^2} = \dfrac{525}{1225} = \dfrac{3}{7}$
i.e. $\;$ $\left(\dfrac{1}{1 + r + r^2}\right) \times \left(\dfrac{1 + r^2 + r^4}{1 + r + r^2}\right) = \dfrac{3}{7}$
i.e. $\;$ $\left(\dfrac{1}{1 + r + r^2}\right) \times \left[r^2 + \dfrac{1 - r^3}{1 + r + r^2}\right] = \dfrac{3}{7}$
i.e. $\;$ $\left(\dfrac{1}{1 + r + r^2}\right) \times \left[r^2 + \dfrac{\left(1 - r\right) \left(1 + r + r^2\right)}{1 + r + r^2}\right] = \dfrac{3}{7}$
i.e. $\;$ $\dfrac{r^2 + 1 - r}{1 + r + r^2} = \dfrac{3}{7}$
i.e. $\;$ $7 r^2 + 7 - 7 r = 3 + 3r + 3 r^2$
i.e. $\;$ $4 r^2 - 10r + 4 = 0$
i.e. $\;$ $2r^2 - 5r + 2 = 0$
i.e. $\;$ $\left(2r - 1\right) \left(r - 2\right) = 0$
i.e. $\;$ $r = \dfrac{1}{2}$ $\;$ or $\;$ $r = 2$
Substituting for $r$ in equation $(1)$ gives
when $\;$ $r = \dfrac{1}{2}$, $\;$ $a = \dfrac{35 \times \dfrac{1}{2}}{1 + \dfrac{1}{2} + \left(\dfrac{1}{2}\right)^2} = \dfrac{35 \times 4}{7 \times 2} = 10$
when $\;$ $r = 2$, $\;$ $a = \dfrac{35 \times 2}{1 + 2 + 2^2} = \dfrac{70}{7} = 10$
$\therefore \;$ The three numbers in G.P when $\;$ $a = 10$, $\;$ $r = \dfrac{1}{2}$ $\;$ are
$\dfrac{10}{1/2}, \; 10, \; 10 \times \dfrac{1}{2}$ $\;\;\;$ i.e. $\;$ $20, \; 10, \; 5$
or when $\;$ $a = 10$, $\;$ $r = 2$ $\;$ are
$\dfrac{10}{2}, \; 10, \; 10 \times 2$ $\;\;\;$ i.e. $\;$ $5, \; 10, \; 20$
$\implies$ The three numbers are $\;$ $5, \; 10, \; 20$