Trying to find a formula to calculate percentage reduction required per month

You want to solve

$$a_0=416.7$$

$$a_1=a_0(1-x)$$

$$a_2=a_1(1-x)$$

$$\cdots$$

$$a_n=a_{n-1}(1-x)$$

and

$$\sum_{n=0}^{11} a_n=4500$$

(we go to $11$ since we started at $0$) for $x$. We shall show that

$$a_n=a_0(1-x)^n$$

is true by induction. It is trivial to prove for $n=0$. Now, assume it is true for $n\geq 0$. Then

$$a_{n+1}=a_n(1-x)=a_0(1-x)^n(1-x)=a_0(1-x)^{n+1}$$

Thus, our sum is

$$4500=416.7\sum_{n=0}^{11}(1-x)^n$$

This is just a finite Geometric Series

$$10.7991=\frac{4500}{416.7}=\sum_{n=0}^{11}(1-x)^n=\frac{1-(1-x)^{12}}{1-(1-x)}=\frac{1-(1-x)^{12}}{x}$$

Now, this is much more difficult to solve as when we expand the right side we get

$$10.7991=-x^{11}+12 x^{10}-66 x^9+220 x^8-495 x^7+792 x^6-924 x^5+792 x^4-495 x^3+220 x^2-66 x+12$$

which is an $11$th degree polynomial which is difficult to solve analytically (probably impossible). However, we do get a value for $x$ which is

$$x=0.0193958=1.93958\%$$

which is indeed very close to $2\%$.