Refinancing a loan. About Suppose someone takes out a home improvement loan for $30,000. The annual interest on the loan is 6% and is compounded monthly. The monthly payment is $600. Let an denote the amount owed at the end of the nth month. The payments start in the first month and are due the last day of every month.
(a) Give a recurrence relation for an. Don't forget the base case.
(b) Suppose that the borrower would like a lower monthly payment. How large does the monthly payment need to be to ensure that the amount owed decreases every month? Feedback?

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MrRoyal MrRoyal · Answer 1 · 2021-03-17T16:21:10+01:00

Answer:

[tex](a)[/tex] [tex]A_n = A_{n-1}(1.005) - 600[/tex] where [tex]A_0 = 30000[/tex]

(b) Above $150

Explanation:

Given

[tex]Loan = \$30000[/tex]

[tex]Rate =6\%[/tex] --- annually

Solving (a): Recursion for the amount at the end of n month

The base case is:

[tex]A(0) = 30000[/tex]

Next, we calculate the monthly rate (r)

[tex]r = \frac{Annual\ Rate}{12}[/tex]

[tex]r = \frac{6\%}{12}[/tex]

[tex]r = 0.5\%[/tex]

[tex]r = 0.005[/tex]

The loan amount remaining at the end of month n is then calculated as:

[tex]A_n = A_{n-1}*(1 + r) - 600[/tex] ---[The 600 represents the monthly payment]

[tex]A_n = A_{n-1}*(1 + 0.005) - 600[/tex]

[tex]A_n = A_{n-1}(1.005) - 600[/tex]

Solving (b):

Suppose the borrower requests for a lower monthly payment, then the following condition will exist:

[tex]P > A_{n-1} *0.005[/tex]

i.e. the monthly payment will exceed the monthly interest

Let [tex]n= 1[/tex]

[tex]P > A_{1-1} *0.005[/tex]

[tex]P > A_0 *0.005[/tex]

Substitute 30000 for [tex]A_0[/tex]

[tex]P > 30000 *0.005[/tex]

[tex]P > 150[/tex]

His monthly payment must exceed $150