The sum of the first two terms of a geometric progression (G.P) is $15$. The first term exceeds the common ratio of the progression by $\dfrac{25}{3}$. Find the fourth term of the progression.
Let the first term of the G.P be $= t_1 = a$
and the common ratio be $= r$
$n^{th}$ term of G.P $= t_n = a \cdot r^{n - 1}$
$\therefore \;$ Second term of G.P $= t_2 = a \cdot r$
Given: $\;$ $t_1 + t_2 = 15$
i.e. $\;$ $a + a \cdot r = 15$
i.e. $\;$ $a \left(1 + r\right) = 15$ $\;\;\; \cdots \; (1)$
And: $\;$ $t_1 = r + \dfrac{25}{3}$
i.e. $\;$ $a = r + \dfrac{25}{3}$ $\;\;\; \cdots \; (2)$
$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes
$\left(r + \dfrac{25}{3}\right) \times \left(1 + r\right) = 15$
i.e. $\;$ $r^2 + \dfrac{28 r}{3} + \dfrac{25}{3} = 15$
i.e. $\;$ $3r^2 + 28 r - 20 = 0$
i.e. $\;$ $3 r^2 + 30 r - 2 r - 20 = 0$
i.e. $\;$ $3r \left(r + 10\right) - 2 \left(r + 10\right) = 0$
i.e. $\;$ $\left(r + 10\right) \left(3r - 2\right) = 0$
i.e. $\;$ $r = -10$ $\;$ or $\;$ $r = \dfrac{2}{3}$
Substituting the value of $r$ in equation $(2)$ gives
when $\;$ $r = -10$, $\;$ $a = -10 + \dfrac{25}{3} = \dfrac{-5}{3}$
and when $\;$ $r = \dfrac{2}{3}$, $\;$ $a = \dfrac{2}{3} + \dfrac{25}{3} = 9$
Now, fourth term of the G.P $= t_4 = a \cdot r^3$
$\therefore \;$ when $\;$ $a = \dfrac{-5}{3}$, $\;$ $r = -10$, $\;$ $t_4 = \dfrac{-5}{3} \times \left(-10\right)^3 = \dfrac{5000}{3}$
and when $\;$ $a = 9$, $\;$ $r = \dfrac{2}{3}$, $\;$ $t_4 = 9 \times \left(\dfrac{2}{3}\right)^3 = \dfrac{8}{3}$