The fourth term of a geometric progression (G.P) exceeds the second term by $24$ and the sum of the second and the third terms is $6$. Find 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 $= t_n = a \times r^{n - 1}$
$\therefore \;$ $2^{nd}$ term of G.P $= t_2 = a \cdot r$
$3^{rd}$ term of G.P $= t_3 = a \cdot r^2$
$4^{th}$ term of G.P $= t_4 = a \cdot r^3$
Given: $\;$ $t_4 = t_2 + 24$
i.e. $\;$ $a \cdot r^3 = a \cdot r + 24$
i.e. $\;$ $a \cdot r \cdot \left(r^2 - 1\right) = 24$
i.e. $\;$ $a \cdot r \cdot \left(r + 1\right) \left(r - 1\right) = 24$
i.e. $\;$ $a \cdot r \cdot \left(r + 1\right) = \dfrac{24}{r - 1}$ $\; \; \; \cdots \; (1)$
And: $\;$ $t_2 + t_3 = 6$
i.e. $\;$ $a \cdot r + a \cdot r^2 = 6$
i.e. $\;$ $a \cdot r \cdot \left(r + 1\right) = 6$
i.e. $\;$ $\dfrac{24}{r - 1} = 6$ $\;\;\;$ [in view of equation $(1)$]
i.e. $\;$ $r - 1 = 4$ $\implies$ $r = 5$
Substituting the value of $r$ in equation $(1)$ gives
$5 \times a \times \left(5 + 1\right) = \dfrac{24}{5 - 1}$
i.e. $\;$ $5 \times 6 \times a = 6$
$\implies$ $a = \dfrac{1}{5}$
$\therefore \;$ The G.P is $\;\;$ $a, \; a \cdot r, \; a \cdot r^2, \; a \cdot r^3, \cdots$
i.e. $\;$ $\dfrac{1}{5}, \; \dfrac{1}{5} \times 5, \; \dfrac{1}{5} \times 5^2, \; \dfrac{1}{5} \times 5^3, \cdots$
i.e. $\;$ $\dfrac{1}{5}, \; 1, \; 5, \; 25, \cdots$