The sum of the first three terms of a $G.P.$ is $16$ and the sum of the next three terms is $128$. Determine the first term,the common ratio,and the sum to $n$ terms of the $G.P.$

(N/A) Let the $G.P.$ be $a, ar, ar^{2}, ar^{3}, ar^{4}, ar^{5}, \dots$
According to the given condition:
$a + ar + ar^{2} = 16$ --- $(1)$
$ar^{3} + ar^{4} + ar^{5} = 128$ --- $(2)$
From $(1)$,$a(1 + r + r^{2}) = 16$
From $(2)$,$ar^{3}(1 + r + r^{2}) = 128$
Dividing $(2)$ by $(1)$,we get:
$\frac{ar^{3}(1 + r + r^{2})}{a(1 + r + r^{2})} = \frac{128}{16}$
$r^{3} = 8 \Rightarrow r = 2$
Substituting $r = 2$ in $(1)$:
$a(1 + 2 + 4) = 16$ $\Rightarrow 7a = 16$ $\Rightarrow a = \frac{16}{7}$
The sum to $n$ terms $S_{n}$ is given by $S_{n} = \frac{a(r^{n} - 1)}{r - 1}$
$S_{n} = \frac{16}{7} \times \frac{(2^{n} - 1)}{2 - 1} = \frac{16}{7}(2^{n} - 1)$