To calculate Christopher's monthly mortgage payment, we will use the following mortgage payment formula:

$M = \frac{P \cdot r \cdot (1 + r)^n}{(1 + r)^n - 1}$

Where:

• $M$ is the monthly payment.
• $P$ is the loan principal, which is $250,000.
• $r$ is the monthly interest rate (annual interest rate divided by 12).
• $n$ is the total number of monthly payments (loan term in years multiplied by 12).

Step-by-Step Breakdown:

1. Loan principal ($P$):
   $P = 250,000$

2. Annual interest rate:
   The annual interest rate is 5.7%. To find the monthly interest rate, divide this by 12: $r = \frac{5.7\%}{12} = \frac{0.057}{12} = 0.00475$

3. Number of monthly payments ($n$):
   Since the loan is taken over 15 years, multiply the number of years by 12: $n = 15 \times 12 = 180 \text{ months}$

4. Substitute values into the formula: $M = \frac{250,000 \times 0.00475 \times (1 + 0.00475)^{180}}{(1 + 0.00475)^{180} - 1}$

Now let's compute the value of $M$.

Calculation:

$M = \frac{250,000 \times 0.00475 \times (1.00475)^{180}}{(1.00475)^{180} - 1}$

Using a calculator to evaluate the exponential terms:

$(1.00475)^{180} \approx 2.359$

Now plug this back into the equation:

$M = \frac{250,000 \times 0.00475 \times 2.359}{2.359 - 1}$ $M = \frac{250,000 \times 0.00475 \times 2.359}{1.359}$ $M \approx \frac{2,801.56}{1.359} \approx 2,061.11$

So, Christopher's monthly mortgage payment is approximately $2,061.

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Would you like more details on any step or have any questions?

Relative Questions:

1. How does compounding frequency affect loan payments?
2. What would Christopher's payment be if the interest rate were different?
3. How much interest will Christopher pay over the life of the loan?
4. What if the loan term is extended to 30 years—how does this change the payment?
5. How can Christopher reduce his monthly payment without changing the loan amount?

Tip:

The shorter the loan term, the higher the monthly payment, but the total interest paid over the loan's life decreases significantly.