To model your situation (in a ${continuous}$ fashion) you can use $y'=(a+bt)y$ which has a solution of $y(t)=y(0)e^{at+bt^2/2}$. To adjust $a$ and $b$ use
$\displaystyle \ln{y(t)\over y(0)}=at+bt^2/2$.
In your case $a=0.1$ and $t=3$ and the targeted growth is $y(3)/y(0)=2$. So $\ln(2)=0.1*3+b*3^2/2$
giving $b=(\ln(2)-0.1*3)*(2/3^2)=0.087$.
This again is for a continuous model. Companies act in step-wise/discrete fashion. In that case you have to use $y'=ay$ for one year, then $y'=(a+b)y$ for the next year, and then $y'=(a+2b)y$ in the next year, etc.
Considering the equation$$A = M \frac{\left(1+\frac{R}{N}\right)^{NT}-1}{\frac RN}\left(1+\frac{R}{N}\right)\tag1$$ define
$$a=\frac AM \qquad, \qquad x=\frac RN\qquad,\qquad n=NT$$ to make the equation
$$a=\frac{(1+x)^n-1} x(1+x) \tag 2$$ What we know is that $x\ll 1$; so, let us develop the rhs using the binomial theorem
$$a=n+\sum_{k=1}^\infty \binom{n+1}{k+1}x^k$$ Transform it as simple Padé approximants which could be
$$a=n\frac{ (n+5) x+6}{6-2 (n-1) x}\implies x=\frac{6 (a-n)}{2 a (n-1)+n (n+5)}$$
Let us try with $a=500$ and $n=200$. The above formula would give
$$a=\frac{3}{400}=0.00750$$ while the exact solution would be $0.00809$; this is not so bad taking into account the huge value assigned to $a$.
Better, but at the price of a quadratic equation
$$a=n\frac{60 +6 (n+13) x+ (n^2+3 n+20) x^2 } {60-24 (n-2) x+3 (n-2) (n-1) x^2 }$$ For the worked example, selecting the "reasonable" root, this would give $x=0.00812$
For sure, we could continue improving but this would be at the price of cubic or quartic equations which can be solved with radicals. To give you an idea, using cubic equations, we should get $x=0.00809422$ while the exact solution is $0.00809450$.
Play with that and, please, tell me how it works for your cases.
Edit
The first way described above is "neutral" in the sense that $a$ is set equal to the ratio of polynomials leading to linear, quadratic, cubic or quartic equations in $x$.
There is another way. Rewrite $(2)$ as $(3)$
$$\frac 1a=\frac{x}{(x+1) \left((x+1)^n-1\right)}\tag 3$$ Expand the rhs as a Taylor series centered at $x=0$ and use series reversion to get
$$x=t+\frac {b_1} 6 t^2+\frac {b_2} {36} t^3+\frac {b_3} {1080} t^4+\frac {b_4} {6480} t^5+\frac {b_5} {90720} t^6+\frac {b_6} {2721600} t^7+O(t^{8})$$ where $t=\frac{2(a-n)}{(n+1)a}$. The coefficients are listed in the table below
$$\left(
\begin{array}{cc}
k & b_k \\
1 & n+5 \\
2 & 2 n^2+11 n+23 \\
3 & 22 n^3+153 n^2+402 n+503 \\
4 & 52 n^4+428 n^3+1437 n^2+2438 n+2125 \\
5 & 300 n^5+2836 n^4+11381 n^3+24879 n^2+30911 n+20413 \\
6 & 3824 n^6+40692 n^5+188712 n^4+496259 n^3+799917 n^2+780417 n+411779
\end{array}
\right)$$
Defining
$$x_{(p)}=t+\sum_{k=0}^p \frac{b_k}{c_k}\,t^{k+1}$$ for the worked example we should get the following values
$$\left(
\begin{array}{cc}
p & x_{(p)} \\
0 & 0.0059701493 \\
1 & 0.0071879409 \\
2 & 0.0076739522 \\
3 & 0.0078882746 \\
4 & 0.0079897311 \\
5 & 0.0080399577 \\
6 & 0.0080655906 \\
7 & 0.0080789649 \\
\cdots & \cdots \\
\infty &0.0080945103
\end{array}
\right)$$
Best Answer
The cummulative total is $S_n=100+110+121$. It can be transformed by factoring out 1.1 and $1.1^2$.
$S_n=100+1.1\cdot 100+1.1^2\cdot 100$ This is the partial sum of a geometric series.
The formula for the partial sum of a geometric series is $S_n=C_0\cdot \frac{(1+g)^n-1}{g}$
$C_0$ is the starting population. $g$ is the growth rate. In your example $C_0=100,n=3$ and $S_n=100$
The formula can be transformed. $S_n\cdot g=C_0\cdot (1+g)^n-C_0$ This equation has to be solved for g.
If n=2, then you have a quadratic equation. It easy to solve.
In your case it is n=3: Here you can apply Cardano´s method or use an approximation method like the Newton–Raphson method. Both methods are time-consuming.