Find repayment amount on loan repaying a partial amount of the principal in a definite period of time. I'm trying to calculate the Monthly payment amount on a loan where I want to be able to define the Principal amount that should be repaid and an Outstanding balance after the repayments. 
Example: 
1749€ is the initial principal amount supposed to be paid in 24 months but I want to take 486,44€ of it and repay that amount in 6 months with a 10% APR compounded monthly. So at the end of the 6 months period, I should have an outstanding balance of 1262,66€. 
the amount should be 93,96€ but can't find the perfect formula for it.
Thanks!
 A: I'm not sure I'm interpreting your question correctly. Let me know.
You have a loan for 1749 with annual interest (APR) at 10% compounded monthly. You want to make monthly payments for six months so that the outstanding balance will be 1262,66 at the end.
The formula is this
$$P = \dfrac{r}{(1+r)^n -1}(PV(1+r)^n - OB)$$
where 


*

*$r$ is in the interest rate

*$n$ is the number of periods (months)

*$PV$ is the initial loan amount

*$OB$ is the outstanding balance you desire 


So with $r=0,10/12$, $n=6$, $PV=1749$, and $OB=1262,66$ we can solve for $P$ and find
$P=93,96$. So in order to have an outstanding balance of $1262,66$ at the end of the six months, you need to pay $93,96$ a month or a total amount of $93,96*6=563,76$.
Addendum: I hope I can provide you with some intuition so this isn't just a formula. Hopefully you know
$$PV(1+r)^n$$
gives the total amount due with interest. So 
$$PV(1+r)^n - OB$$
is the total amount you'll have to pay in order for $OB$ to be left. Say you pay $P$ each period. You take out the loan on January 1. You make your first payment on February 1. When the 5 months have gone by, that first payment will have $P(1+r)^5$ value. On March 1 you make your second payment $P$. When 4 of the remaining months have gone by that second payment will have $P(1+r)^4$ value. Continue this for April 1, May 1, and so on. Let's call $TP$ the total payment. Hopefully it is clear that
$$TP = P + P(1+r)^1 + P(1+r)^2 + P(1+r)^3 + P(1+r)^4 + P(1+r)^5$$
This can seem very magical, but bear with me. We can simplify this with a clever trick. Multiply by (1+r) and subtract to get
\begin{align*}
    TP(1+r) &= P(1+r) + P(1+r)^2 + P(1+r)^3 + P(1+r)^4 + P(1+r)^5 + P(1+r)^6\\\\   
    -\phantom{y=}&\\    
    TP &= P + P(1+r)^1 + P(1+r)^2 + P(1+r)^3 + P(1+r)^4 + P(1+r)^5\\
   \_\_\_\_\_\_\_\_\_\_&\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\
    TP(1+r) - TP &= P(1+r)^6 - P    
\end{align*}
Now we can solve for $TP$
\begin{align}
TP((1+r) - 1) &= P((1+r)^6 - 1 )\\
TP &= P\dfrac{(1+r)^6 - 1 }{(1+r) - 1)}\\
&=P\dfrac{(1+r)^6 - 1}{r}
\end{align}
Now we can find the payment amount by setting total payment equal to total balance minus left over balance.
\begin{align}
TP &= PV(1+r)^6 - OB\\
P\dfrac{(1+r)^6 - 1}{r} &= PV(1+r)^6 - OB\\
P &= \dfrac{r}{(1+r)^6 - 1}(PV(1+r)^6 - OB)
\end{align}
This was for $n=6$ but you can extrapolate to general $n$. I hope this provides some insight.
