The sum of the first and the third term of a geometric progression (G.P) is $20$ and the sum of the first three terms is $26$. Find the progression.
Let the first term of the G.P $= t_1 = a$ $\;$ and the common ratio $= r$
$n^{th}$ term of G.P $= t_n = a \cdot r^{n - 1}$
$\therefore \;$ Second term of G.P $= t_2 = a \cdot r$
Third term of G.P $= t_3 = a \cdot r^2$
Given: $\;$ $t_1 + t_3 = 20$
i.e. $\;$ $a + a \cdot r^2 = 20$
i.e. $\;$ $a \left(1 + r^2\right) = 20$ $\implies$ $a = \dfrac{20}{1 + r^2}$ $\;\;\; \cdots \; (1)$
And: $\;$ $t_1 + t_2 + t_3 = 26$
i.e. $\;$ $a + a \cdot r + a \cdot r^2 = 26$
i.e. $\;$ $a \left(1 + r + r^2\right) = 26$
i.e. $\;$ $\left(\dfrac{20}{1 + r^2}\right) \times \left(1 + r + r^2\right) = 26$ $\;\;\;$ [In view of equation $(1)$]
i.e. $\;$ $\dfrac{1 + r + r^2}{1 + r^2} = \dfrac{26}{20} = \dfrac{13}{10}$
i.e. $\;$ $10 + 10 r + 10 r^2 = 13 + 13 r^2$
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 $\;$ $r = \dfrac{1}{3}$
When $r = 3$, we have from equation $(1)$, $a = \dfrac{20}{1 + 3^2} = 2$
When $r = \dfrac{1}{3}$, we have from equation $(1)$, $a = \dfrac{20}{1 + \left(\dfrac{1}{3}\right)^2} = 18$
$\therefore \;$ The required G.P is
when $a = 2, \; r = 3$: $\;\;\;$ $2, \; 6, \; 18, \; 54, \cdots$
when $a = 18, \; r = \dfrac{1}{3}$: $\;\;\;$ $18, \; 6, \; 2, \; \dfrac{2}{3}, \cdots$