I'm trying to solve these two equations: $$12\times{}y = \sum_{m=0}^{11}\sum_{n=0}^{m}r\times(1+r-r\times{}t)^n$$ $$12\times{}y\times{}s = \sum_{m=0}^{11}\sum_{n=0}^{m}r\times{}t\times{}(1+r-r\times{}t)^n$$
$y$ and $s$ are known. $r$ and $t$ are unknown. I do not know how to proceed. I cannot see a way to simplify or isolate the unknowns. Does anyone have an idea how to proceed?
First, note that if you divide the LHS of the second equation by the LHS of the first, you just get $s$. If you divide the RHS of the second equation by the RHS of the first, you just get $t$. Since the two sides are equal in both equations, the ratios are both of the same two numbers, and therefore are the same. That is, $t = s$. (Note that this doesn't work when $y=0$, because then you would be dividing by $0$. When $y = 0$, $t$ can have any value.)
To solve for $r$ we use GNU Supporter's comment. The sum of a finite geometric series is $$\sum_{i=0}^k a^i = \frac{a^{k+1} - 1}{a - 1}$$
But to simplify the notation a bit, let $u = 1 - t = 1- s$, then the first equation becomes
$$12y = \sum_{m=0}^{11}\sum_{n=0}^mr(1 + ru)^n$$
Now $$\sum_{n=0}^mr(1 + ru)^n = r\frac{(1 + ru)^{m+1} - 1}{ru} = \frac 1u((1 + r - tr)^{m+1} - 1)$$ and $$\begin{align}\sum_{m=0}^{11}\sum_{n=0}^mr(1 + ru)^n &= \frac 1u\sum_{m=0}^{11}\left((1 + ru)^{m+1}-1\right)\\&=\frac 1u\left(\sum_{m=1}^{12}(1 + ru)^{m}-12\right)\\&=\frac 1u\left(\sum_{m=0}^{12}(1 + ru)^{m}-13\right)\\&=\frac1u\left(\frac{(1 + ru)^{13} - 1}{ru}-13\right)\\&=\frac{(1 + ru)^{13} - 1}{ru^2}-\frac{13}{u}\end{align}$$
Therefore $$12y = \frac{(1 + ru)^{13} - 1}{ru^2}-\frac{13}{u}\\12yru^2 = (1 + ru)^{13} -13ru$$ which is a 13-degree polynomial in $r$. You can simplify it a bit by making the substitution $v = 1 + ru$ and noting that $ru = 1-v$, so $$12yu(1-v) = v^{13} - 13(1-v)\\v^{13} + (13+12yu)v -(13+12yu) = 0$$ But that is the best you can do exactly. There is no nice "quadratic formula"-style way of expressing the roots of a 13-degree polynomial, so you would have to solve this by approximation techniques such as Newton's method. I can tell you that it has 1 to 3 real roots. The remaining roots will all be complex numbers.
Once you do obtain a root for $v$, you can recover $$r = \frac{1 - v}u = \frac{1-v}{1-s}$$