[BruceZ]

[ QUOTE ]
If someone loans you $10,000 at 4% interest, and you are to pay 400/mo, how many months are you going to pay. At 0 its 25 months. Assume the normal parameters that a car loan company would use, like how its compounded or whatever. Thanks and be well and happy.

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I'm sure there are calculators all over the place which do this, and financial types probably know the formula, but let's see if we can derive it from first principles.

Let P = the principal = $10,000
Let i = the monthly interest = 0.04/12
Let m = the monthly payment = $400
And n = the number of months to pay off loan

After the 1st month, our balance will have grown to P*(1 + i) and we will pay m, so our balance will be P*(1 + i) - m. We compute the new balance at the end each month as the previous month's balance times (1 + i) minus m:

month 2: [P*(1 + i) - m]*(1 + i) - m
= P*(1 + i)^2 - m*(1 + i) - m

month 3: [P*(1 + i)^2 - m*(1 + i) - m]*(1 + i) - m
= P*(1 + i)^3 - m*(1 + i)^2 - m*(1 + i) - m

...

month n: P*(1 + i)^n - m*sum{k = 0 to n-1} (1 + i)^k = 0

We set this equal to zero since after n months the loan will be paid off. Now we solve for n. First we sum the geometric series as sum{k = 0 to n-1} x^k = (1 - x^n) / (1 - x). See derivation of this at the bottom. [img]/images/graemlins/diamond.gif[/img]

P*(1 + i)^n = m*[1 - (1 + i)^n] / -i

-P*i/m * (1 + i)^n = 1 - (1 + i)^n

(1 - P*i/m)*(1 + i)^n = 1

Taking logs of both sides:

ln(1 - P*i/m) + n*ln(1 + i) = 0

n = -ln(1 - P*i/m) / ln(1 + i)

n = -ln[1 - 10,000*(0.04/12)/400] / ln(1 + .04/12)

n =~ 26.15 months This means we make 26 monthly payments of $400 and one final payment of $58.76. Paying it off in exactly 26 months would require monthly payments of $402.17.


[img]/images/graemlins/diamond.gif[/img] Derivation of geometric series used above:

S = sum{k = 0 to n-1} x^k

S = 1 + x + x^2 + x^3 + ... + x^(n-1)

x*S = x + x^2 + x^3 + ... + x^(n-1) + x^n
----------------------------------------------------------
S - x*S = 1 - x^n

S = (1 - x^n) / (1 - x) [img]/images/graemlins/diamond.gif[/img]