Find the first term and the common ratio of a geometric progression if the sum of its first three terms is $10.5$ and the difference between the first and the fourth term is equal to $31.5$.
Let the first term of the G.P be $= t_1 = a$ $\;$ and common ratio $= r$
$n^{th}$ term of G.P $= t_n = a \cdot r^{n-1}$
$\therefore \;$ $2^{nd}$ term $= t_2 = a \cdot r$, $\;$ $3^{rd}$ term $= t_3 = a \cdot ar^2$, $\;$ $4^{th}$ term $= t_4 = a \cdot r^3$
Given: $\;$ $t_1 + t_2 + t_3 = 10.5$
i.e. $\;$ $a + a \cdot r + a \cdot r^2 = 10.5$
i.e. $\;$ $a \left(1 + r + r^2\right) = 10.5$ $\;\;\; \cdots \; (1)$
And: $\;$ $t_4 - t_1 = 31.5$
i.e. $\;$ $a \cdot r^3 - a = 31.5$
i.e. $\;$ $a \left(r^3 - 1\right) = 31.5$
i.e. $\;$ $a = \dfrac{31.5}{r^3 - 1}$ $\;\;\; \cdots \; (2)$
$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes
$\left[\dfrac{31.5}{r^3 - 1}\right] \times \left(1 + r + r^2\right) = 10.5$
i.e. $\;$ $\left[\dfrac{31.5}{\left(r - 1\right) \left(r^2 + r + 1\right)}\right] \times \left(1 + r + r^2\right) = 10.5$
i.e. $\;$ $\dfrac{31.5}{r - 1} = 10.5$
i.e. $\;$ $31.5 = 10.5 r - 10.5$
i.e. $\;$ $10.5 r = 42$ $\implies$ $r = \dfrac{42}{10.5} = 4$
Substituting the value of $r$ in equation $(2)$ gives
$a = \dfrac{31.5}{4^3 - 1} = \dfrac{31.5}{63} = 0.5$
$\therefore \;$ The first term of the given G.P is $= a = 0.5$ and its common ratio is $= r = 4$