



 
Hi Kenneth, First of all you need to know how to calculate the sum of a finite geometric series. If you have a geometric sequence, $a, at , at^2, at^3, \cdot, \cdot, \cdot$ and $S_n$ is the sum of the first $n$ terms of this sequence then \[S_n = a + at + at^2 + at^3 + \cdot, \cdot \cdot + at^{n1} = \frac{a \left( 1  t^n\right)} {1t}.\] A proof appears in Penny's response to a previous question. I have changed Penny's notation as you have already used the variable $r.$ We will use this result later. You have a loan amount of $P$ dollars nd an interest rate of $p$ per period. At the end of each period you make a payment of $A$ dollars. At the end of the first period the amount you owe has grown to $P(1+r)$ dollars and you pay $A$ dollars so at the beginning of the second period you owe \[P(1+r)  A \mbox{ dollars.}\] At the end of the second period the amount you owe is $(P(1+r)  A)(1+r)$ dollars and you pay $A$ dollars so at the beginning of the third period you owe \[(P(1+r)  A)(1+r) A = P(1+r)^2  A\left( (1+r) + 1 \right) \mbox{ dollars.}\] Continuing in this fashion at the end of the third period you owe \[P(1+r)^3  A\left( (1 + r)^2 + (1+r) + 1 \right) \mbox{ dollars}\] and at the end of the $n^{th}$ period you owe \[P(1+r)^n  A\left( (1+r)^{n1} + \cdot \cdot \cdot + (1 + r)^2 + (1+r) + 1 \right) \mbox{ dollars}\] But at the end of the $n^{th}$ period you have paid off the loan so \[P(1+r)^n  A\left( (1+r)^{n1} + \cdot \cdot \cdot + (1 + r)^2 + (1+r) + 1 \right) = 0.\] At this point you ned to use the expression for the sum of the first $n$ terms of a geometric sequence given above with $t = 1+r$ which gives \[P(1+r)^n  A\left( \frac{1  (1+r)^n}{1  (1+r)} \right) = 0.\] Solving this equation for $A$ gives \[A = P \left( \frac{r(1+r)^n}{(1+r)^n  1}\right) \mbox{ dollars.}\] Harley 



Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences. 