Question

use $\text{pmt} = \frac{p\left(\frac{r}{n}\
ight)}{\left1-\left(1+\frac{r}{n}\
ight)^{-nt}\
ight}$ to determine the regular payment amount, rounded to the nearest cent. the cost of a home is financed with a $130,000 30-year fixed-rate mortgage at 4.5%.
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.)

Explanation:

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Part a:

Step1: Define variables

$P = 130000$, $r = 0.045$, $n = 12$, $t = 30$

Step2: Calculate monthly rate & periods

$\frac{r}{n} = \frac{0.045}{12} = 0.00375$; $-nt = -12*30 = -360$

Step3: Compute denominator

$1 - (1+0.00375)^{-360} \approx 1 - 0.2607 = 0.7393$

Step4: Calculate monthly payment

$PMT = \frac{130000*0.00375}{0.7393} \approx 662.63$

Step5: Total payments & interest

Total payments: $662.63*360 = 238546.80$; Interest: $238546.80 - 130000 = 108546.80$

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Part b:

Step1: Month 1 calculations

Interest: $130000*0.00375 = 487.50$; Principal: $662.63 - 487.50 = 175.13$; Remaining balance: $130000 - 175.13 = 129824.87$

Step2: Month 2 calculations

Interest: $129824.87*0.00375 \approx 486.84$; Principal: $662.63 - 486.84 = 175.79$; Remaining balance: $129824.87 - 175.79 = 129649.08$

Step3: Month 3 calculations

Interest: $129649.08*0.00375 \approx 486.18$; Principal: $662.63 - 486.18 = 176.45$; Remaining balance: $129649.08 - 176.45 = 129472.63$

Answer:

a. Monthly payment: $\$662.63$; Total interest: $\$108,546.80$
b.

Payment Number	Monthly Payment	Interest Paid	Principal Paid	Remaining Balance
2	$\$662.63$	$\$486.84$	$\$175.79$	$\$129,649.08$
3	$\$662.63$	$\$486.18$	$\$176.45$	$\$129,472.63$