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I know the simple interest formula is, $I = prt$, but I don't know the interest to be able to use that formula.

Here's the variables. Please help me with the formula.

$$p = \$92,000.00$$

$$t = 72 \text{ months}$$

$$ \mathrm{Pmt} = \$1461.00$$

$$r = ?$$

This is a simple interest loan paid monthly and the rate is a fixed annual percentage.

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  • $\begingroup$ You must be more precise. We're just guessing at what you might mean. What's the interest scheme? How frequent are the payments? Is this a loan with life $t$? More information, please. $\endgroup$ – MPW Nov 6 '15 at 3:06
  • $\begingroup$ Thank you MPW, I updated the question with more details. $\endgroup$ – SirOracle Nov 6 '15 at 3:12
  • $\begingroup$ This question didn't have anything to do with (boolean-algebra)... $\endgroup$ – Math1000 Nov 6 '15 at 3:33
  • $\begingroup$ I didn't know what tag to put on it but it required one and I'm not a math whiz..... Sorry $\endgroup$ – SirOracle Nov 6 '15 at 3:35
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I suppose that you need to use the classical formula $$A = P\frac{i(1 + i)^n}{(1 + i)^n - 1}$$ where $A$ is the monthly payment amount, $P$ the amount of the loan, $i$ the monthly interest and $n$ the total number of payments.

Everything is simple to compute using this formula except the interest (otherwise you would not ask the question) and, as far as I know, only numerical methods would be able to solve the problem.

Since $i$ is small compared to $1$, we can use Taylor expansion around $i=0$ and get as an approximation $$A=\frac{P}{n}+\frac{ (n+1) P}{2 n}i+\frac{ \left(n^2-1\right) P}{12 n}i^2+O\left(i^3\right)$$ which reduces to quadratic equation in $i$. Applied to your numbers $(A=1461$, $P=92000$, $n=72)$, the positive root is given by $$i=\frac{4 \sqrt{1175358765}-125925}{2980225}\approx 0.00376113$$ Using this number would lead to $A=1460.98$ which is not bad.

A rigorous calculation, based on Newton method, would lead to $i\approx 0.00376161$.

You can notice that, neglecting the second order term of the expansion, we would have obtained $$i=\frac{1649}{419750}\approx 0.00392853$$ which is too high but already quite accurate.

If you want to stay simple, you could use the simplest Pade approximant and write $$A=P \frac{\frac{i (n+2)}{3 n}+\frac{1}{n}}{\frac{1}{6} i (1-n)+1}$$ For your numbers, this would give $$i=\frac{9894}{2635579}\approx 0.00375401$$

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  • $\begingroup$ These formulas are for compound interest, are they not? I think the OP was asking about simple interest. $\endgroup$ – David K Nov 6 '15 at 14:35
  • $\begingroup$ @DavidK. May I confess that I don't know anything in the financial area, interest and, thanks to that, I guess that the bank blesses me everyday ! $\endgroup$ – Claude Leibovici Nov 7 '15 at 3:11
  • $\begingroup$ For what it's worth, I don't think I've ever seen simple interest except maybe in a textbook or a homework problem. The formulas you gave are the ones that I think would apply in real life. $\endgroup$ – David K Nov 7 '15 at 4:41

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