A loan of £12,000 is issued and is repaid in instalments of £300 at the end of each month for 4 years. Calculate the effective annual rate of interest for this loan.
What I tried. The equation of value is $$ 12,000=3,600\,a_{\overline{4}|}^{(12)} $$
But how to solve this equation for $i$?
Please help.
The equation is
$$\large{12,000\cdot (1+\frac{i}{12})^{48}=300\cdot \frac{\left(1+ \frac{i}{12} \right)^{48}-1}{\frac{i}{12}}}$$.
Substituting $1+\frac{i}{12}$ by $x$.
$12,000\cdot x^{48}=300\cdot \frac{x^{48}-1}{x-1} \qquad |\cdot (x-1)$
$12,000\cdot x^{49}-12,000\cdot x^{48}=300\cdot x^{48}-300$
$12,000\cdot x^{49}-12,300\cdot x^{48}+300=0$
This is a 49th degree polynomial. It cannot be solved algebraically. Therefore you have to use an approximation method, for instance the Newton–Raphson method.
For a loan of $12,000$ and payments $P=300$ at monthly interest rate $i_m$ for $n=48$ months, you have to solve $$ L=P\,a_{\overline{n}|i_m}\quad \Longrightarrow\quad \frac{1-(1+i_m)^{-48}}{i_m}=40 $$ and you can solve it numerically and find $i_m\approx 0.77\%$. Thus the effective annual rate of interest $i$ is $$ i=(1+i_m)^{12}-1\approx 9.64\% $$
