The APR for a loan with constant repayments made at regular intervals can be calculated by solving this formula:
http://www.financeformulas.net/Loan_Payment_Formula.html
s = 1000;
n = 52;
t = 60/52;
The periodic payments amount to a total cost of $2,000.
pp = 2000.0/60;
This next step solves pp = (s p)/(1 - (1 + p)^(-n t))
for p
. (Mathematica used.)
p = Last@Reduce[pp == (s p)/(1 - (1 + p)^(-n t)), p, Reals]
0.0263204
Calculating the effective annual interest rate, r
, from the periodic rate, p
:
r = (1 + p)^52 - 1
2.86112
The effective annual rate is 286.11 %
ref: http://en.wikipedia.org/wiki/Effective_interest_rate
or, if a nominal rate, i
, is required:
i = p * 52 = 0.0263204 * 52 = 136.9 % nominal rate compounded weekly
Calculation Check
Checking the principal is calculated correctly. This is basically the summation in the page linked in the OP's question:
http://en.wikipedia.org/wiki/Annual_percentage_rate#European_Union
i.e. s = Σ pp (1 + r)^-(k/n)
for k = 1 to 60
The loan formula used earlier is actually inducted from the summation.
So, running the check by back-calculating the loan principal:
Sum[pp (1 + r)^-(k/n), {k, 60}]
1000
Yup, checks out.
Note the summation uses the effective annual rate, not the nominal rate.