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Let's say a loan of £1000 was to be repaid by payments of £100 at the end of each quarter, and a smaller final payment made one quarter after the last regular payment. If the interest rate is 10% p.a, but it drops to 8% p.a once the outstanding balance is less than £500, then how would I calculate which payment causes the outstanding balance to drop to less than £500?
I know that in theory this could be done by drawing up a loan schedule, but how could I approach this question algebraically?
Hint: The outstanding balance after $n$ quarterly made payments is
$$B_n=1000\cdot \left( 1+\frac{0.1}4 \right)^n-100\cdot \frac{\left( 1+\frac{0.1}4 \right)^n-1}{\frac{0.1}4 }$$
Therefore the outstanding balance is less than 500, if $B_n<500$. You have to solve it for $n$. If we multiply the inequality by $\frac{0.1}4 =\frac1{40}$ and substitute $1+\frac{0.1}4$ by $q$ we obtain
$$25\cdot q^n-100\cdot \left( q^n-1\right)<12.5$$
Remark:
In this context I can only refer to problems where a loan ($C_0$) is taken out and the repayments are constant. If the repayments are made $m$ times in a year and the (constant) repayments (r) are made at the end of each period we have the following term:
$$B_n=C_0\cdot \left( 1+\frac{i}m \right)^{n}-r\cdot \frac{\left( 1+\frac{i}m \right)^{n}-1}{\frac{i}m }$$
$B_n$: Outstanding balance after $n$ periods.
$i$: nominal interest rate
Usually we have a constant interest rate i and a constant repayment (annuity). In this case we can set $B_n$ equal to zero and we are able to calculate the missing value of $i, n,r$ or $C_0$
If the repayments are made at the $\color{blue}{\textrm{beginning}}$ of each period the formula changes to
$$B_n=C_0\cdot \left( 1+\frac{i}m \right)^{n}-r\cdot \color{blue}{ \left( 1+\frac{i}m \right)}\cdot \frac{\left( 1+\frac{i}m \right)^{n}-1}{\frac{i}m }$$
