I would be extremely grateful if someone could help with this problem. I'm planning a series of lessons on basic finance and wanted to brush up. I will not have to teach this but I like to keep a step ahead of students.

Suppose I wish to borrow £3000 with an Annual Interest Rate of 24.9% over 60 months. A set up fee of £25 is charged ( I want to consider the two cases - one where the set up fee is incorporated into the amount borrowed and one it separately added up front)

I'm struggling to calculate the APR in either case.

I know it represents the annual cost of the borrowing including fees but how exactly is it calculated?
I have looked at calculator soup which has an APR figure of 25.3159% but I just can't replicate on my own. I have managed to reproduce the monthly repayment amounts (essentially using geometric series) and interest but not APR.

My best effort is to get an APR of 25.73% This was found by computing $\frac{£3000 \times (1.249)+£25}{£3000}-1$

Here is the calculator soup website.


In truth I'm not sure whether it's my maths that's dodgy or a lack of understanding of finance. I would especially appreciate answers that give clear reasons and not just formula. The website only uses the information I've provided but I'm at a loss.

Edit 2 I have decided to use the formula $\frac{a(1+a)^N}{(1+a)^N-1}=\frac{P}{C}$ shown on http://www.efunda.com/formulae/finance/apr_calculator.cfm and just teach trial and improvement for $a$. I'd still be interested in seeing a simple derivation though. In particular what are both sides of the equation.

Once Again thanks to anyone that has considered the question.

  • $\begingroup$ The wikipedia page for APRwill show you that there's no universally agreed upon definition of the term. If you need to teach this perhaps start with one definition where you can compute the answer and be sure to tell your students that the result may differ when they negotiate their own particular loan. $\endgroup$ Jan 10, 2016 at 22:38
  • $\begingroup$ Thanks. Need to keep it simple $\endgroup$
    – Karl
    Jan 10, 2016 at 22:40
  • $\begingroup$ May be, this could be interestiing : math.stackexchange.com/questions/1536653/… $\endgroup$ Jan 12, 2016 at 12:04
  • $\begingroup$ @ClaudeLeibovici Thanks I will look at that pen and paper calculation. $\endgroup$
    – Karl
    Jan 12, 2016 at 12:24

1 Answer 1


Let P be the monthly amount that you need to pay for the total amount of Borrowing which is the loan amount C and the added cost, say, E ( fees especially). As a borrower you need to consider C+E as the total amount borrowed and find the monthly payment by treating the Borrowed funds C+E. Let us say you pay monthly an interest rate of "r". Then you will have to find the monthly payments discounted over the entire period that will equal the present value which is the borrowed funds C+E.

And that is :

$$C+E = \frac{P}{1+r} + \frac{P}{(1+r)^2} + \frac{P}{(1+r)^3} + \cdots + \frac{P}{(1+r)^N}$$

$$C+E = \frac{P}{1+r}\left(1+\frac{1}{1+r}+ \frac{P}{(1+r)^2}+\cdots+\frac{P}{(1+r)^{N-1}}\right)$$

$$C+E = \frac{P}{1+r}\left(\dfrac{\left(1-\frac{1}{(1+r)^N}\right)}{\left(1-\frac{1}{(1+r)}\right)}\right)$$

If you simplify this, you get

$$C+E = \dfrac{P((1+r)^N-1)}{r(1+r)^N}$$

Now rearraging, we get

$$P = (C+E)\dfrac{r(1+r)^N}{(1+r)^N-1}$$.

Using this monthly instalment and now the the loan amount "alone" you find the interest rate "a" that equals the monthly payment. i.e

$$P =C\dfrac{a(1+a)^N}{(1+a)^N-1}$$

$$\frac{P}{C} = \dfrac{a(1+a)^N}{(1+a)^N-1}$$

$$ \dfrac{a(1+a)^N}{(1+a)^N-1} - \frac{P}{C} = 0$$.

This can only be found iteratively by numerical methods ( you could use Newton Raphson method, Secant method or Binary search method) and find "a"

This "a" is called the APR according to the loan Advertisement.

  • $\begingroup$ Thanks for this detailed answer. I'm going to carefully ready and try to understand.(+1) $\endgroup$
    – Karl
    Jan 12, 2016 at 11:21
  • $\begingroup$ @karl, if you find the answer useful, could you accept by clicking on the tick mark. $\endgroup$ Jan 12, 2016 at 11:26
  • $\begingroup$ I can almost follow the whole thing which is beautifully clear The only thing I'm stuck on is the first line It's the set up. I will give it more thought and will accept the answer once I've got my head around it. $\endgroup$
    – Karl
    Jan 12, 2016 at 11:54

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