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If a mortgage principal is $200,000, interest rate is 3.50% and the payment is $900 a month. Is there an equation to find how many payments it would take to pay off the loan?

Edit 1

I have tried to solve n and have not been able too. So I tried just solving P to see if your equation comes out with a 1000 and it does not appear to.

I wrote your equation in Python to solve P as shown below. Maybe you can tell me what I am doing wrong?

P=1000
R=50
i=0.04
n=20
print (1-(1+i/12)**n)/(i/12)*R
#The above prints out -1032.30906873

Thank you for your help.

Community
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null_pointer
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  • In Edit 1, you wrote (1+0.04/12)**n for the formula $(1+i)^{-n}.$ The part where you write 0.04/12 is OK, because the $i$ in $(1+i)^{-n}$ is the interest per payment period (which is once a month in your example). But notice the minus sign in the exponent of $(1+i)^{-n}$: you should have written **(-n) instead of **n. – David K May 11 '17 at 18:00
  • My fault. So now if I do print (1-(1+i/12)**-n)/(i/12)*R it comes out to 965.839416178. The problem is I don't know how to solve n. I tried a few different ways using Logarithm but its jut not working. – null_pointer May 11 '17 at 18:36

2 Answers2

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Yes, $P=(\frac{1-(1+i)^{-n}}{i})R$. R is how much you pay per period (\$ 900), P is your principal (\$ 200000), and i is your interest rate per period ($\frac{0.035}{12}$). Now solve for n to compute how many months it takes to pay this mortgage off.

Sean Nemetz
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  • Ok, how do I solve n above? Can you by any chance give me an example? – null_pointer May 10 '17 at 11:38
  • Say you wanted to know how long it would take to pay off a loan of $$ 1000$ at an interest rate of 4% compounded monthly paying $$ 50$ monthly. Plugging into the equation you have that $$1000=\frac{1-(1+\frac{0.04}{12})^{-n}}{\frac{0.04}{12}}*(50).$$ Solving for $n:$ – Sean Nemetz May 10 '17 at 22:11
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So I found the answer to my question here. So someone can mark this as a duplicate.

Anyway, thank you @Sean Nemetz for trying to help.

null_pointer
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