Calculating monthly loan repayments Apologies if this is a too trivial question but I'm teaching myself and can't get my answer to match the one in my text book.
The task is to calculate the monthly repayments of a £500 loan to be repaid in two years. Interest on the remaining debt is calculated monthly and charged at 11% p.a. First repayment a month after loan given.
Here's my attempt: 
First I figured the monthly interest charge,  M, as
$$M = 1.11^\frac {1}{12}$$
After the first month, if a repayment of $\chi$ is made the remaining debt would be
$$ 500M - \chi $$
After two months
$$ (500M - \chi)M - \chi   =  $$
$$500M^2 - \chi M - \chi $$
After n months
$$ 500M^n - \chi M^{n-1} - \chi M^{n-2} ... \chi M^1 - \chi$$
Or
$$ 500M^n - \frac{\chi (M^n - 1)}{M - 1} $$
I reckon this should equal zero after 24 repayments so, rearranging
$$ \chi = \frac{500M^{24} (M - 1)}{M^{24} - 1} $$
which comes to  £23.18 but the answer given is £23.31. I've tried different numbers of charges/payments and the nearest I got was 
$$ \chi = \frac{500M^{25} (M - 1)}{M^{24} - 1} $$
equalling  £23.38
Can anyone see where I'm going wrong? I guess it could be a typographical error but it'd be the only one I've spotted  (so far.) 
Here's the question exactly as stated in case I'm missing something there 
A bank loan of £500 is arranged to be repaid in two years by equal monthly instalments. Interest, calculated monthly, is charged at 11% p.a. on the remaining debt. Calculate the monthly repayment if the first repayment is to be made one month after the loan is granted. 
 A: Let's look at this in general. We need to pay someone $x$ money units over $n$ payments with an interest rate of $r>1$ after each payment. We want to pay an equal amount, say $y$ money units, every payment. Let $x_i$ denote the amount of money we need to pay after $i$ payments. Then we find
$$
x_0=x
\\x_1=r(x_0-y)=rx-ry
\\x_2=r(x_1-y)=r^2x-(r^2+r)y
\\...
\\x_i=r^ix-(r^i+r^{i-1}+...+r)y.
$$
This gets quite messy. However, there is a formula for $r^i+r^{i-1}+...+r$. To derive this, let's say that $S=r^i+r^{i-1}+...+r$. Then $rS=r^{i+1}+r^i+r^{i-1}+...+r^2$, so $rS+r=r^{i+1}+S$. Hence $r-r^{i+1}=S-rS=(1-r)S$, so we find $S=\frac{r-r^{i+1}}{1-r}$. Plugging this into $x_i$ we find
$$x_i=r^ix-\frac{r-r^{i+1}}{1-r}y.$$
We want to be done with paying after $n$ payments, so we need to find $y$ such that $x_n=0$. So we finally just need to do some algebra.
$$
0=x_n=r^nx-\frac{r-r^{n+1}}{1-r}y
\\r^nx=\frac{r-r^{n+1}}{1-r}y
\\y=\frac{1-r}{r-r^{n+1}}r^nx.
$$
In your case $x=500$ and $n=24$ and $r=1.11^{\frac1{12}}$, so you can just plug it into the formula.
A: If the interest is payed m times a year then you usually use the period interest rate $i_m=\frac{i}{m}$. With $m=12$ and $i=0.11$ we get $i_{12}=\frac{0.11}{12}$ Therefore the equation is
$$500\cdot \left(1+\frac{0.11}{12} \right)^{24}=x\cdot \frac{\left(1+\frac{0.11}{12} \right)^{24}-1}{\frac{0.11}{12} }$$
$$x=500\cdot \left(1+\frac{0.11}{12} \right)^{24}\cdot \frac{0.11}{12\cdot \left(\left(1+\frac{0.11}{12} \right)^{24}-1\right)}=23.30391\approx 23.30$$
