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

\[S_n = a + at + at^2 + at^3 + \cdot, \cdot \cdot + at^{n-1} = \frac{a \left( 1 - t^n\right)} {1-t}.\]

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

\[P(1+r)^n - A\left( (1+r)^{n-1} + \cdot \cdot \cdot + (1 + r)^2 + (1+r) + 1 \right) \mbox{ dollars}\]

But at the end of the $n^{th}$ period you have paid off the loan so

\[P(1+r)^n - A\left( (1+r)^{n-1} + \cdot \cdot \cdot + (1 + r)^2 + (1+r) + 1 \right) = 0.\]

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