Given a geometric sequence $U(n)$, its only known values are $U_{0}$, $S$ the sum of the geometric sequence till a certain $N$-th term.
Is it possible to find $q$ via the following sum equation: $$ S = U_{0}\left(\frac{q^N - 1}{q - 1}\right)\;? $$ To simplify what I'm trying to find, I have set the following variables: $$ X = \frac{S}{U_{0}}. $$ What I want to know is that given $X$ and $N$, can we find $q$ easily with the following equation: $$ X = \frac{q^N - 1}{q-1}\;? $$
If $n \ge 5$, in general, there is no solution in radicals for $q$ in the equation $$x = \frac{q^n-1}{q-1}. \tag{1}$$ For smaller $n$, we can use the methods for solving low-order polynomials (e.g., quadratic, cubic, and quartic formulae).
However, we can quickly find a root of $(1)$ by employing Newton's method. Let $$f(z) = z^n - 1 - x(z-1) = z^n - xz + (x-1). \tag{2}$$ Then $$f'(z) = nz^{n-1} - x,$$ and we have the recursion $$z_{k+1} = z_k - \frac{f(z_k)}{f'(z_k)} = z_k - \frac{z_k^n - xz_k + (x-1)}{nz^{n-1} - x} = \frac{(n-1)z_k^n - (x-1)}{nz_k^{n-1} - x}. \tag{3}$$ We now need to make an initial guess $z_0$. A natural choice is to suppose $x \approx q^n/q = q^{n-1}$, so that $$z_0 = x^{1/(n-1)}. \tag{4}$$ (We will assume $n > 1$ since the case $n = 1$ is trivial.) Then we compute sucessive iterates of $z_k$ until we obtain convergence.
Here is an example. Suppose $x = 3725$ and $n = 13$. Then we choose $z_0 = 3725^{1/12} \approx 1.98424$ and we use $(3)$ to compute the following table:
$$\begin{array}{c|c} k & z_k \\ \hline 0 & 1.98423842903653327007689553768 \\ 1 & 1.90092746706785318059143692471 \\ 2 & 1.86622449798215768509689978419 \\ 3 & 1.86093724741670620970910598569 \\ 4 & 1.86082688465257555395503309422 \\ 5 & 1.86082683752952357276321384119 \\ 6 & 1.86082683752951498527167126885 \\ 7 & 1.86082683752951498527167126856 \\ 8 & 1.86082683752951498527167126856 \\ \end{array}$$
Therefore $q \approx 1.86082683752951498527167126856$. Now we test it: $$\frac{(1.8608268)^{13} - 1}{1.8608268 - 1} \approx 3725.$$