Find the APR at which total interest paid exceeds principal So I had an assignment to calculate details about a mortgage and I did that just fine. One of the things that caught my curiosity was that the interest rate offered, 3.247% APR, did not produce a total interest paid that exceeded the principal. I thought in fact that it generally did.
The formula I believe I should be using is: 
$P*\frac{i*(1+i)^n}{(1+i)^n-1}$
At what APR would the bank collect as much in interest paid as the principal loaned? I can see that it is roughly 5.3%, but I want to find it exactly.
I was looking at a 30 year fixed rate on 74900 (with 20% downpayment, so the principal was 59920). I would like to accept either a specific answer or a general one that I can apply, either will do nicely.
 A: The payment and interest scale with principal, so we may as well take the principal to be $1$ and the total of payments $2$.  This gives $$360\frac i{1-(1+i)^{-360}}=2$$  Which Alpha solves with $i \approx 0.004420$ and an annual rate of $$5.304\%$$  You can't solve this equation algebraically for the interest rate.  It has to be done numerically.
A: Assume that the principal is $\$59, 920$.
Let the monthly interest rate be $j$, so that $(1+j)^{n} = 1 + i$, where $i$ is (what we will assume) is the APR. [This is not necessarily going to be the case.]
Suppose that we make monthly payments of $M$.
At each month $t = 1, 2, \dots, 360$, we have $I_t = jB_{t-1}$, $I_t$ being the interest portion of the payment, and $B_{t-1}$ denoting the previous balance. Each payment consists of a principal portion and interest portion, for which $M_t = M = I_t + P_t$.
We wish to sum
$$\sum_{t=1}^{360}I_t = j\sum_{t=1}^{360}B_{t-1}\text{.}$$
Consider $B_0$: this is  $59,920$.
$B_1$ is given by the present value of future payments at interest rate $j$: this is going to be given by $$M\left[\dfrac{1-(1+j)^{-359}}{j} \right]\text{.}$$
Repeating this process, we get
$$Mj\sum_{t=0}^{359}\dfrac{1-(1+j)^{-(360-j)}}{j}$$
as the total interest paid. We wish to solve 
$$Mj\sum_{t=0}^{359}\dfrac{1-(1+j)^{-(360-j)}}{j} > 59,920\text{.}$$
Notice we may simplify the left side:
$$Mj\sum_{t=0}^{359}\dfrac{1-(1+j)^{-(360-j)}}{j} = M\left[360-\sum_{t=0}^{359}(1+j)^{-(360-j)}\right]$$
and the series 
$$\sum_{t=0}^{359}(1+j)^{-(360-j)} = \dfrac{1-(1+j)^{-360}}{j/(1+j)}$$
so our final equation is
$$M\left[360-\dfrac{1-(1+j)^{-360}}{j/(1+j)}\right] > 59,920$$
and this needs to be solved numerically. 
This, furthermore, shows that we need to be given $M$ in order to solve for $j$.
A: at i = 5.72%
30 year term
principal = 59,920
term = 30
$$payment = 59920 / 12 / ((1 - (1 + i)^{30}) / (1 - 1 / (1 + i)))$$
$$ = 59920 / 12 / 15 = $$
$$=332.90pm$$
