How can I solve for r?

Question

How can this be solved: Using the Geometric Series Sum formula: $S_n = \frac{a (r^n-1)}{r-1}$ It is impossible to get $r$ with normal algebra. But with other methods it is possible. for example to find $a$ it would look like this with normal algebra: $a = (Sn*r-Sn)/((r^n)-1)$ So how do i find $r$ ?

Answer 1

If you divide $(1-r)$ into $(1-r^n)$ you have the polynomial $1 + r + r^2 + \cdots + r^{n-1}$. Then you are left with the equation $ar^{n-1} + \cdots + ar + (a-S_n) = 0$. If $r$ is rational then it can be found with the rational root theorem. If not you would have to solve it using an approximation method like Newton's method.

Examples by request:

Example 1. Suppose someone gives you $a=2,~ S_n=\frac{259}{108},~ n=4$ and we wish to determine $r$.

Then we have the equation $$2r^3 + 2r^2 + 2r - \frac{43}{108} = 0 \\ 216r^3 + 216r^2 + 216r -43 = 0$$ If this has a rational root, it it must have a numerator as one of the divisors of the constant term, $\pm1,~ \pm 43$, and a denominator as one of the divisors of the leading coefficient, $1,~ 2,~ 3,~ 4,~ 6,~ 8,~ 9,~ 12,~ 18,~ 24,~ 27,~ 36,~ 54,~ 72,~ 108,~ 216$. Trying these in turn we find $r=\frac{1}{6}$.

Example 2. Suppose someone gives you $a = \sqrt{2}, S_n = 123.2,$ and $n=10$. Because of the uncertainty in the representation by computer of $a$ and possible truncation of $S_n$ we cannot be sure of an exact solution.

We have the equation

$$\sqrt{2}r^9 + \cdots + \sqrt{2}r - 121.78578 = 0$$ We make some guess, $r_0 = 5$. Then we get our new guess $r_1$ by

$$r_{n+1} = r_n + \frac{f(r_n)}{f'(r_n)},~ f'(r) = 9\sqrt{2}r^8 + 8\sqrt{2}r^7 + \cdots + \sqrt{2}$$

After 13 iterations we have an estimate $r=1.4454$ which differs from the previous estimate by less than $0.0001$. Checking this answer in our original formula we get $S_n^* = 123.1952$ which is accurate to what our user provided us with within the precision they gave us.