Sequences Help

dhthompson

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Nov 13, 2009
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Phil bought a car by taking out a loan for $24,500 at 0.5% interest per month. Phil's normal monthly payment is $534.47 per month, but he decides that he can afford to pay $150 extra toward the balance each month. Write a recursively defined sequence for the balance he has each month.
 
He decides to make monthly repayments of 684.47 dollars.\displaystyle He\ decides\ to\ make\ monthly\ repayments\ of\ 684.47\ dollars.

Balance after first month is 24,500(1.005)684.47 dollars.\displaystyle Balance\ after\ first\ month\ is\ 24,500(1.005)-684.47\ dollars.

Balance after second month is [24,500(1.005)684.47](1.005)684.47 dollars\displaystyle Balance\ after\ second\ month\ is\ [24,500(1.005)-684.47](1.005)-684.47\ dollars
=(previous months balance)(1.005)684.47 dollars\displaystyle =(previous\ month's\ balance)(1.005)-684.47\ dollars
=24,500(1.005)2684.47(1.005)684.47 dollars.\displaystyle =24,500(1.005)^2-684.47(1.005)-684.47\ dollars.

Balance after third month is [24,500(1.005)2684.47(1.005)684.47](1.005)684.47\displaystyle Balance\ after\ third\ month\ is\ [24,500(1.005)^2-684.47(1.005)-684.47](1.005)-684.47
=(previous months balance)(1.005)684.47 dollars.\displaystyle =(previous\ month's\ balance)(1.005)-684.47\ dollars.
=24,500(1.005)3684.47(1.005)2684.47(1.005)684.47 dollars.\displaystyle =24,500(1.005)^3-684.47(1.005)^2-684.47(1.005)-684.47\ dollars.

Try formulating from here.\displaystyle Try\ formulating\ from\ here.
 
or:

a = 24500, p = 684.47, i = .005, n = number of months

After n months:

a(1+i)^n - p[(1+i)^n - 1] / i
 
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