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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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