Trying to find a formula to calculate percentage reduction required per month I've been messing around with compound interest calculations, but can't quite get this one right.
I have a starting January value: $416.7$
I would like to reduce that value by the same percent "P" each month throughout the year from Jan thru Dec, so that when the amounts for each month add up, they equal $4500$.
For example, if the percentage is $2\%$:
Jan: 416.7
Feb: 408.3 (Feb= Jan-(Jan*0.02))
Mar: 400.2 (Mar= Feb-(Feb*0.02))
...
Total of all months=4500

$2\%$ is close (I get $4485$), but I need to calculate what that exact percent needs to be, since the starting value and the total will change in my scenario.
What would the correct formula be to find that percentage?
 A: 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 $2\%$.
A: List the information we have:


*

*starting value: $a_o = 416.7$

*percentage: $p$

*sum of differences: $4500$
Now we can find a recursive expression for the value at each month in terms of $a_0$, $p$ and $n$.
\begin{align}
\text{Jan:} && a_o & = 416.7 \\
\text{Feb:} && a_1 & = a_o - p * a_0 \\
\text{Mar:} && a_2 & = a_1 - p * a_1 \\
&& ... \\
&& a_n & = a_{n-1}\left ( 1 - p \right )
\end{align}
What we want to do next is sum all of the $a_n$'s for the months of January until December, that is $n \in [0,11]$, and this should equal 4500.
\begin{equation}
\sum^n_0 a_n = 4500
\end{equation}
To be able to sum this easily, we'd like to have an expression in terms of $n$. You can brute force this expression (write out $a_1$, $a_2$, $a_3$, ... in terms of $a_0$ and you'll see the pattern) or know that this is a Geometric series. 
\begin{equation}
a_n = a_0 (1-p)^n
\end{equation}
You can now use the equation for the sum of the first $n$ terms of a Geometric series to find $p$ (check the link).
