For 1≤n≤10, find a formula for pn, the payment in year n on a loan of 100000 dollars. Interest is 3 percent per year, compounded annually, and payments are made at the end of each year for 10 years. Each payment is 10000 dollars plus the interest on the amount of money outstanding.
p(n)=

Question

05-11-2019
Mathematics

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For 1≤n≤10, find a formula for pn, the payment in year n on a loan of 100000 dollars. Interest is 3 percent per year, compounded annually, and payments are made at the end of each year for 10 years. Each payment is 10000 dollars plus the interest on the amount of money outstanding.
p(n)=

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sammyayol2013 sammyayol2013 · Answer 1 · 2019-11-05T09:21:24+01:00

Answer: Pn= 10000(1+ 0.03(11-n))

Step-by-step explanation:

From the question it is given that

Pn= 10000 + 3% of Dn eqn 1

Where Pn is payment at year n

And Dn is outstanding debt at year n

n= 1,2,3...,10

D1= 100000

D2=100000-10000

Dn=100000-10000(n-1) eqn2

Therefore, substituting eqn2 into eqn1

Pn= 10000+3% of (100000-10000(n-1))

Pn= 10000+ 0.03(100000 - 10000(n-1))

Pn= 10000(1+ 0.03(10 - (n-1)))

Pn= 10000(1+ 0.03(11-n))

MrRoyal MrRoyal · Answer 2 · 2021-10-07T11:07:46+02:00

The question is an illustration of annual compound interest.

The function of each payment is: [tex]\mathbf{P(n) = 13300 + 300n }[/tex]

The given parameter are:

[tex]\mathbf{Amount = 100000}[/tex]

[tex]\mathbf{Payment = 10000 + Rate \times Outstanding}[/tex]

[tex]\mathbf{Rate = 3\%}[/tex]

Represent the outstanding debt with D.

So, we have:

[tex]\mathbf{D_1 = 100000 - 10000 \times (1 - 1)}[/tex]

[tex]\mathbf{D_2 = 100000 - 10000 \times (2 - 1)}[/tex]

[tex]\mathbf{D_3 = 100000 - 10000 \times (3 - 1)}[/tex]

For the nth year, the outstanding debt would be

[tex]\mathbf{D_n = 100000 - 10000 \times (n - 1)}[/tex]

Recall that:

[tex]\mathbf{Payment = 10000 + Rate \times Outstanding}[/tex]

So, the equation becomes

[tex]\mathbf{Payment = 10000 + Rate \times (100000 - 10000 \times (n - 1))}[/tex]

Substitute [tex]\mathbf{Rate = 3\%}[/tex]

[tex]\mathbf{Payment = 10000 + 3\% \times (100000 - 10000 \times (n - 1))}[/tex]

[tex]\mathbf{Payment = 10000 + 3000 - 300 \times (n - 1)}[/tex]

[tex]\mathbf{Payment = 13000 - 300 \times (n - 1)}[/tex]

Open bracket

[tex]\mathbf{Payment = 13000 - 300n + 300}[/tex]

Collect like terms

[tex]\mathbf{Payment = 13000 + 300 - 300n }[/tex]

[tex]\mathbf{Payment = 13300 + 300n }[/tex]

Express as a function

[tex]\mathbf{P(n) = 13300 + 300n }[/tex]

Hence, the required function is:

[tex]\mathbf{P(n) = 13300 + 300n }[/tex]

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For 1≤n≤10, find a formula for pn, the payment in year n on a loan of 100000 dollars. Interest is 3 percent per year, compounded annually, and payments are made at the end of each year for 10 years. Each payment is 10000 dollars plus the interest on the amount of money outstanding. p(n)=

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For 1≤n≤10, find a formula for pn, the payment in year n on a loan of 100000 dollars. Interest is 3 percent per year, compounded annually, and payments are made at the end of each year for 10 years. Each payment is 10000 dollars plus the interest on the amount of money outstanding.
p(n)=