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# Practice Paper 3 Question 12

A bank has a loan scheme with an annual interest rate of $$0<r<1$$ ($$100r$$ is the annual percentage rate). The bank charges interest every month. You take a loan of $$L$$ pounds and wish to pay it back in exactly $$m$$ months by making monthly payments of $$p$$ pounds each. Find $$p$$.

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## Warm-up Questions

1. Evaluate the sum $$a+ar+ar^2+\cdots+ar^n,$$ with $$r \neq 0$$ and $$n>0.$$
2. Given $$a_0= 5$$ and $$a_n= 3a_{n-1}+4,$$ find $$a_4.$$

## Hints

• Hint 1
Find an expression for the amount remaining to be paid after month $$i.$$
• Hint 2
... in terms of the amount remaining to be paid in month $$i-1.$$
• Hint 3
You might first want to find the cumulative monthly interest rate. Hint: it's not $$r/12.$$
• Hint 4
... knowing that the cumulative monthly interest rate is the interest rate that gives $$r$$ when applied successively $$12$$ times.
• Hint 5
If $$x_i$$ denotes the amount remaining to be paid after month $$i,$$ you should now have a recursive formula for $$x_i.$$ Try to find the explicit formula.
• Hint 6
You should get an expression that contains $$x_0.$$ What is the value of $$x_0?$$

## Solution

Let $$x_i$$ be the amount remaining to be paid after month $$i$$. Thus, $$x_0=L.$$ We have $$x_i=x_{i-1}(1+\mu)-p,$$ where $$\mu$$ is the cumulative monthly interest rate (we'll determine its value later). To simplify our working, let $$c=1+\mu.$$ Unwinding the recursion, we get: \begin{align} x_i &= x_{i-1}c-p \\ &= (x_{i-2}c-p)c - p \\ &= \left(\left((x_{i-3}c-p)c - p\right) \right)c-p \\ &= \ldots \\ &= x_0c^i-p\sum_{k=0}^{i-1}c^k \\ &= Lc^i-p\frac{c^i-1}{c-1}. \end{align}

We want $$x_m=0,$$ which gives $$p=L\;c^m\;\frac{c-1}{c^m-1}.$$

Now we need to express $$\mu$$ in terms of $$r.$$ The cumulative monthly interest rate $$\mu$$ is the interest rate that gives $$r$$ when applied successively (i.e. recursively) $$12$$ times. This is not equal to $$r/12\;^{(*)}.$$ An initial amount $$z_0$$ becomes $$z_0(1+\mu)$$ after a month, and $$z_0(1+\mu)^{12}$$ after $$12$$ months, which must also be equal to $$z_0(1+r).$$ This is in fact another recursion: $$z_i=z_{i-1}(1+\mu),$$ where we have $$z_{12}=z_0(1+\mu)^{12}=z_0(1+r).$$ Hence, $$\mu=(1+r)^{1/12}-1.$$

The monthly payment sought is then $$p=L\;(1+r)^{m/12}\;\frac{(1+r)^{1/12}-1}{(1+r)^{m/12}-1}.$$

(*) $$\mu$$ can in fact be approximated by $$r/12$$ for $$r\ll1$$ though Taylor Series. This is a safe assumption for savings accounts, but not for loans (hmm, what does that tell you about banks?).

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