Solving $ \frac{r^{200}-1}{r^{199}-1} = \frac{\alpha}{\beta}$ in a question on Geometric Progressions Question:

Let $a_n$ be the $n^{th}$ term of a geometric progression of positive numbers. Let $$\sum_{n=1}^{200} a_n = \alpha$$ and let $$\sum_{n=1}^{199} a_n = \beta$$ such that $\alpha \neq \beta$. Then find the common ratio of the GP.

My solution:
$a_1 + a_2 + a_3 + ... + a_{200} = \alpha$
$a_1 + a_2 + a_3 + ... + a_{199} = \beta$
$\implies a_{200} = \alpha - \beta = a_1r^{199}$
Now,
$a_1(1+r+r^2+r^3+...+r^{199}) = \alpha$
$a_1(1+r+r^2+r^3+...+r^{198}) = \beta$
$$\implies \frac{a_1(1-r^{200})}{1-r} = \alpha, \frac{a_1(1-r^{199})}{1-r} = \beta$$
$$\implies \frac{r^{200}-1}{r^{199}-1} = \frac{\alpha}{\beta}$$
After this step I am not able to find $r$ in terms of $\alpha$ and $\beta.$
I tried everything from cross-multiplying to trying to expand the terms(by reducing powers), componendo-dividendo etc. Please suggest me a way to solve it.
Any help would be highly appreciated!
 A: You have $\alpha-r\beta=a_1$ and as you found, $\alpha-\beta=a_1r^{199}=(\alpha-r\beta)r^{199}$
$$\implies \beta r^{200}-\alpha r^{199}+(\alpha-\beta)=0$$
Let $\gamma = \frac\alpha\beta$, then we have $f(r)=r^{200}-\gamma r^{199}+(\gamma-1)=0$.  This is a polynomial in $r$ with two sign changes ($\gamma>1$ for positive numbers), so by the rule of Descartes, there are $0$ or $2$ positive roots. There are no negative roots.
$r=1$ is always a root of that polynomial, so there must be exactly two positive real roots for the polynomial.  However if $r=1$ is to be the common ratio of the GP, then $\gamma=\frac{200}{199}=\gamma_0$ should also be true from the original equations for consistency, which makes it a double root (why? check the sign of $f'(1)$) and the GP is essentially a constant progression.
If $\gamma \neq \gamma_0$, then there is always one more positive root which is larger than $1$ (if $\gamma >\gamma_0$) and less than $1$ (if $1<\gamma < \gamma_0$) [again the sign of $f'(1)$ is enough to show this].  This root is then the common ratio of the GP, however to find this one you would need to use numerical methods.
For a quick approximation when $\gamma > \gamma_0$ i.e $r > 1$, so $r^{199}$ gets very large,
$$ \gamma-r=\frac{\alpha-\beta}{r^{199}}\to 0 \implies r \to \gamma =\frac{\alpha}{\beta}$$
this approximation gets better very fast as $r$ or $\gamma$ gets large. Of course, when $\gamma< \gamma_0$, ie $r<1$, this approximation doesn't work.
A: Notice that $$\frac {r^{200}-1}{r^{199}-1}\approx r$$
for values of $r$ even slightly greater than $1$. For example,  $$\frac {1.01^{200}-1}{1.01^{199}-1}\approx 1.0116$$
This approximation only becomes more and more accurate as $r$ increases. Once you cross $r=1.05$, you might as well change the $\approx$ to a $=$. This is why a reasonable approximation is $r=\frac \alpha\beta $ for $\frac \alpha\beta>1.05$.
