Remember to register **here** for FREE to ask any questions you may come across in your QCE studies!In the real world, you probably won't just take out money from the bank, and pay a lump sum of that + a **** ton of interest in 32 years time. On one hand, the banks will get angry at you. On the other, your debt just increases more.

Reducing balance loans are what people usually end up doing. That is, they pay off the loan in instalments (bit by bit). For our course, we make life easy by assuming that each payment is of equal value.

The recursive model given on your formula sheet is

\[ A_{n+1}=rA_n -R. \]

It's derived under the scenario that "after the last payment, the amount owing (\(A_n\)) will gain interest between the two periods (\(r\)). Then at the end of this period (and hence the start of the next), a payment (\(R\)) gets made before we calculate the new amount owing (\(A_{n+1}\)),

As a quick example, suppose we take out an initial loan value of \(\$10000\), and have to pay interest at \(12\%\) p.a. compounded quarterly (4 periods a year). Then we can set \(A_0 = 10000\), and \(r = 1+\frac{0.12}{4} = 1.03\) - note that the compounding frequency had to be catered for here! Then we set \(A_n\) to be the balance after \(n\)

**quarter**-years.

To add to the example, let's suppose the quarterly payment is $750. Then the balance after 1 quarter-year will be

\[ A_1 = 1.03\times 10000 - 750 = 9550.\]

We can continue doing this. The balance after 2 quarter-years will be

\[ A_2 = 1.03\times 9550 - 750 =9086.50 \]

And just one more for now:

\[A_3 = 1.03\times 9086.50 - 750 =8609.095 \]

It's clear that the balance is indeed going down. If we repeat this process over and over, at some value of \(n\), the value of \(A_n\) will be less than \(R\). Then, we only have to pay that bit off to settle the loan.

Harder problems don't necessarily do this though - they might omit something. They might ask you to physically calculate the monthly repayment, the number of periods or something else. There are usually (but NOT always) better approaches to these, such as using the annuity formula \( M\left( \frac{1-(1+i)^{-n}}{i} \right)\) to handle these cases.