`m=(((r)/(1,200))(1+(r)/(1,200))^(N))/((1+(r)/(1,200))^(N)-1)P`
The formula above gives the monthly payment m needed to pay off a loan of P dollars at r percent annual interest over N months. Which of the following gives P in terms of m, r, and N ?

Options

A. `P=((1+(r)/(1,200))^(N)-1)/(((r)/(1,200))(1+(r)/(1,200))^(N)-1)m`
B. `P=((1+(r)/(1,200))^(N)-1)/(((r)/(1,200))(1+(r)/(1,200))^(N))m`
C. `P=((r)/(12,00))m`
D. `P=((1,200)/(r))m`

Answer

To find \( P \) in terms of \( m \), \( r \), and \( N \) from the given formula:

\[ m = \frac{\left(\frac{r}{1200}\right) \left(1 + \frac{r}{1200}\right)^{N}}{\left(1 + \frac{r}{1200}\right)^{N} - 1} P \]

we will follow these steps:

### Step 1: Multiply both sides by the denominator
Multiply both sides of the equation by \( \left(1 + \frac{r}{1200}\right)^{N} - 1 \) to eliminate the fraction on the right side:

\[ m \left(1 + \frac{r}{1200}\right)^{N} - m = \frac{r}{1200} \left(1 + \frac{r}{1200}\right)^{N} P \]

### Step 2: Isolate \( P \)
Now, we can isolate \( P \) by dividing both sides by \( \frac{r}{1200} \left(1 + \frac{r}{1200}\right)^{N} \):

\[ P = \frac{m \left(1 + \frac{r}{1200}\right)^{N} - m}{\frac{r}{1200} \left(1 + \frac{r}{1200}\right)^{N}} \]

### Step 3: Simplify the expression
Factor out \( m \) from the numerator:

\[ P = \frac{m \left(\left(1 + \frac{r}{1200}\right)^{N} - 1\right)}{\frac{r}{1200} \left(1 + \frac{r}{1200}\right)^{N}} \]

### Step 4: Final expression for \( P \)
Now, we can express \( P \) more neatly:

\[ P = \frac{1200m \left(\left(1 + \frac{r}{1200}\right)^{N} - 1\right)}{r \left(1 + \frac{r}{1200}\right)^{N}} \]

This is the expression for \( P \) in terms of \( m \), \( r \), and \( N \).