A question about geometric series/progressions Is it possible to find the value of the common ratio $r$ given the first term and the sum to $n$ terms without using a numerical approach and solving analytically?
In other words, can I simplify
$$\text{10} = \text{5}\,\,\,\frac{r^{\text{10} }-1} {r-1}$$
$$10 = 5\frac {r^{10} - 1} {r - 1}\implies 2 = \frac {r^{10} - 1} {r - 1}\implies 2r-2=r^{10} - 1$$
that is to say
$$r^{10}-2r+1=0$$
…
Where do I go from here? Have I done something wrong?
 A: Just a small detail to start
$$S_n=\sum_{m=0}^n a r^m=a \frac{r^{n+1}-1}{r-1}$$ Let $k=\frac {S_n} a$, cross multiply (as you did) and you want to approximate the zero of function
$$f(r)=r^{n+1}-k r+(k-1)$$ The derivatives are
$$f'(r)=(n+1)r^n-k \qquad \text{and} \qquad f''(r)=n(n+1)r^{n-1}~~>0 $$
The first derivative cancels at
$$r_*=\left(\frac{n+1}{k}\right)^{-1/n} \implies f(r_*) <0$$ and this is a minimum by the second derivative test. The solution will be smaller than $r_*$ since $r=1$ ia the trivial solution.
Notice that
$$f(0)=k-1  \qquad \text{and} \qquad f\left(\frac{k-1}{k}\right)=\left(\frac{k-1}{k}\right)^{n+1}~~>0$$
Expanding as series around $r=\frac{k-1}{k}=t$ which is a lower bound, we have
$$f(r)= t^{n+1}+ \left((n+1) t^n-k\right)(r-t)+t^{n+1}\sum_{p=2}^\infty \binom{n+1}{p} t^{-p}(r-t)^p$$
Truncating the summation to some low order and using series reversion
$$r=t+\frac{t^{n+1}}{k-(n+1) t^n}+\frac{n (n+1) t^{3n+1} }{2 \left(k-(n+1) t^n\right)^3}+\cdots$$
For your example where $n=9$ and $k=2$  (that is to to say $t=\frac 12$) this gives
$$r=\frac{38652555}{77228944}=\color{red}{0.5004931182}278$$ while the exact solution is $\color{red}{0.5004931182865}$.
