A loan is to be repaid with level instalments payable at the end of each half-year for $3$ and $\frac{1}{2}$ years, at a nominal rate of interest of $8\%$ converted semiannually. After the $4^{\rm th}$ payment, the outstanding balance loan is $\$5000$. Find the initial amount of the loan.

Can someone please explain to me how I can do this?


Let $i=0.08/2=0.04$ be the semiannual interest, $n=7$ be the number of compounding periods, and $k=4$ be the number of semiannual payments made. It can be shown that the principal $P$ of the loan when the outstanding balance of the loan is $Q=\$5000$ is

$$P = \frac{1-(1+i)^{-n}}{1-(1+i)^{-(n-k)}} Q$$

With the above numbers, the principal of the loan is $P \approx \$10,814.20$.

A way to show this relation is to understand that, if $m$ is the semiannual loan payment, then the outstanding balance of the loan is given by

$$P(1+i)^k - m (1+i)^{k-1}-m (1+i)^{k-2}-\ldots-m(1+i)-m = Q$$

The semiannual payment is given by

$$m=P \frac{i}{1-(1+i)^{-n}}$$

Sum the geometrical series in the first equation and substitute the semiannual payment in the second equation and the stated result follows.


I just wanted to add a note relative to what @Ross states, that you may simply look at this as issuing a new loan. In fact, this is quite correct, and you can see it almost instantly from the above equation, if you rewrite it as

$$P\frac{ i}{1-(1+i)^{-n}} = Q \frac{i}{1-(1+i)^{-(n-k)}}$$

Here, I multiplied both sides by $i$. What this says is that the monthly payment of the theoretical new loan with the outstanding balance and the remaining number of payments is equal to the original monthly payment. The result follows.


Hint: the loan runs $1 \frac 12$ years after the fourth payment. If you think of it as a new loan of $\$5000$ you can find the payment. Then use that payment, the interest rate, and the full term to get the initial balance.

  • $\begingroup$ (+1): I mentioned your view of the solution as an addendum to mine. The whole thing is very pretty, as far as finance and loans go. $\endgroup$ – Ron Gordon Feb 22 '13 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.