Use PMT \( =\frac{P\left(\frac{r}{n}\right)}{\left[1-\left(1+\frac{r}{n}\right)^{-n t}\right]} \) to determine the regular payment amount, rounded to the nearest cent. The cost of home is financed with a \( \$ 110,000 \) 20-year fixed-rate mortgage at \( 3.5 \% \). a. Find the monthly payments and the total interest for the loan. b. Prepare a loan amortization schedule for the first three months of the mortgage. a. The monthly payment is \( \$ \square \). (Do not round until the final answer. Then round to the nearest cent as needed.)

To apply the PMT formula given, we need to define the variables as follows: - \( P = 110,000 \) (loan amount) - \( r = 0.035 \) (annual interest rate) - \( n = 12 \) (number of payments per year) - \( t = 20 \) (loan term in years) Substituting these into the PMT formula: \[ PMT = \frac{110000 \left(\frac{0.035}{12}\right)}{1 - \left(1 + \frac{0.035}{12}\right)^{-12 \cdot 20}} \] Calculating: 1. Calculate \( \frac{r}{n} = \frac{0.035}{12} = 0.00291667 \). 2. Calculate \( n \cdot t = 12 \cdot 20 = 240 \). 3. Calculate \( \left(1 + \frac{0.035}{12}\right)^{-240} = \left(1.00291667\right)^{-240} \approx 0.431905 \). 4. Therefore, \( 1 - 0.431905 \approx 0.568095 \). 5. Finally, compute: \[ PMT = \frac{110000 \cdot 0.00291667}{0.568095} \approx \frac{320.8337}{0.568095} \approx 564.87 \] So, the monthly payment is \( \$564.87 \). To find the total interest: 1. Total payments over 20 years = \( 564.87 \times 240 = 135564.80 \). 2. Total interest = Total payments - Principal = \( 135564.80 - 110000 = 25564.80 \). **Final Summary:** a. The monthly payment is \( \$ 564.87 \) and the total interest is \( \$ 25564.80 \). Now, let's create a loan amortization schedule for the first three months: **Month 1** - Beginning Balance: \( 110,000 \) - Payment: \( 564.87 \) - Interest: \( 110000 \cdot 0.00291667 \approx 320.83 \) - Principal: \( 564.87 - 320.83 \approx 244.04 \) - Ending Balance: \( 110000 - 244.04 \approx 109755.96 \) **Month 2** - Beginning Balance: \( 109755.96 \) - Payment: \( 564.87 \) - Interest: \( 109755.96 \cdot 0.00291667 \approx 320.51 \) - Principal: \( 564.87 - 320.51 \approx 244.36 \) - Ending Balance: \( 109755.96 - 244.36 \approx 109511.60 \) **Month 3** - Beginning Balance: \( 109511.60 \) - Payment: \( 564.87 \) - Interest: \( 109511.60 \cdot 0.00291667 \approx 319.49 \) - Principal: \( 564.87 - 319.49 \approx 245.38 \) - Ending Balance: \( 109511.60 - 245.38 \approx 109266.22 \) Thus, for the first three months, the amortization schedule is as follows: | Month | Payment | Interest | Principal | Ending Balance | |-------|----------|----------|-----------|----------------| | 1 | 564.87 | 320.83 | 244.04 | 109755.96 | | 2 | 564.87 | 320.51 | 244.36 | 109511.60 | | 3 | 564.87 | 319.49 | 245.38 | 109266.22 |