Geometric Progression: Finding the Outstanding Loan Amount after Each Year

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SUMMARY

John's bank loan of $200,000 incurs an annual interest rate of 3%. He pays $1,000 monthly, resulting in a formula for the outstanding loan amount at the end of each year, denoted as Un. The relationship is established as Un = k(Un-1 - 1200), where k is a constant. Additionally, Un can be expressed in the form of a + (1.03^n)b, with a and b as constants to be determined through calculations involving the loan's principal and interest payments.

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Question:
John took a bank loan of $200000 to buy a flat. The bank charges an anual interest rate of 3% on the outstanding loan at the end of each year. John pays $1000 at the beginning of each month until he finishes paying for his loan.
Let Un denote the amount owed by john at the end of the nth year,

(i)show that Un= k(u n-1 - 1200) where k is a constant to be determined.
(ii) Express u n in the form of a +(1.03^n)b, where a and b are constants to be determined.



Homework Equations





The Attempt at a Solution


year total amt owed (before end of year) ''(at the end of year)
1 $200000 200000(1.03)- 12000
2 200000(1.03)-1200 (1.03)^2(200000)-12000(1.03)


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JJYEO325 said:
Question:
John took a bank loan of $200000 to buy a flat. The bank charges an anual interest rate of 3% on the outstanding loan at the end of each year. John pays $1000 at the beginning of each month until he finishes paying for his loan.
Let Un denote the amount owed by john at the end of the nth year,

(i)show that Un= k(u n-1 - 1200) where k is a constant to be determined.
(ii) Express u n in the form of a +(1.03^n)b, where a and b are constants to be determined.



Homework Equations





The Attempt at a Solution


year total amt owed (before end of year) ''(at the end of year)
1 $200000 200000(1.03)- 12000
that last value is incorrect. "The bank charges an anual interest rate of 3% on the outstanding loan at the end of each year." At the end of the first year he will have paid 12000 so the outstanding loan is 200000- 12000= 188000. You should have 1.03(200000- 12000)= 1.03(200000)- 1.03(12000)
2 200000(1.03)-1200 (1.03)^2(200000)-12000(1.03)
At the end of the second year, the outstanding balance will be 1.03(200000)- 1.03(12000)- 12000 so the interest will be 1.03(1.03(200000)- 1.03(12000)- 12000)= 1.03^2(200000)- (1.03^2(12000)+ 1.03(12000)= 1.03^2(200000)- 12000(1.03+ 1.03^2).

At the end of the third year, the outstanding balance will be 1.03^2(200000)- 12000(1.03+ 1.03^2)- 12000= 1.03^2(200000)- 12000(1+ 1.03+ 1.03^2) so the interest will be 1.03(1.03^2(200000)- 12000(1+ 1.03+ 1.03^2)= 1.03^3(200000)- 12000(1.03+ 1.03^2+ 1.03^3).

Get the idea?
 
Thanks.. So the total amount he owe the bank for the first year is Un >? which is 1.03(200000)-(1.03)12000? Sorry I don't seem to understand the question.
 

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