The sum of the first three terms of an increasing geometric progression (G.P) is $13$ and their product is $27$. Calculate the sum of the first five terms of the progression.
Let the first term of the G.P be $= t_1 = \dfrac{a}{r}$,
the second term $= t_2 = a$ $\;$ and the third term $= t_3 = a \cdot r$
where $\;$ $r$ $\;$ is the common ratio of the G.P
Given: $\;$ $t_1 + t_2 + t_3 = 13$
i.e. $\;$ $\dfrac{a}{r} + a + a \cdot r = 13$
i.e. $\;$ $a \left(\dfrac{1}{r} + 1 + r\right) = 13$ $\;\;\; \cdots \; (1)$
And: $\;$ $t_1 \times t_2 \times t_3 = 27$
i.e. $\;$ $\dfrac{a}{r} \times a \times a \cdot r = 27 = 3^3$
i.e. $\;$ $a^3 = 3^3$ $\implies$ $a = 3$
Substituting $a = 3$ in equation $(1)$ gives
$3 \left(\dfrac{1}{r} + 1 + r\right) = 13$
i.e. $\;$ $3 + 3r + 3r^2 = 13r$
i.e. $\;$ $3r^2 - 10r + 3 = 0$
i.e. $\;$ $\left(3r - 1\right) \left(r - 3\right) = 0$
i.e. $\;$ $r = \dfrac{1}{3}$ $\;$ or $\;$ $r = 3$
But since the first three terms of the given G.P are in an increasing order
$\implies$ $r > 1$
$\therefore \;$ $r = \dfrac{1}{3}$ $\;$ is not an acceptable solution.
Now, sum of first $n$ terms of a G.P $= S_n = \dfrac{t_1 \left(r^n - 1\right)}{r - 1}$
i.e. $\;$ $S_n = \dfrac{\dfrac{a}{r} \left(r^n - 1\right)}{r - 1}$
$\therefore \;$ Sum of first $5$ terms of G.P
$= S_5 = \dfrac{\dfrac{3}{3} \left(3^5 - 1\right)}{3 - 1} = \dfrac{1 \times 242}{2} = 121$