Seth repays a 30-year loan with a payment at the end of each year. Each of the first 20 payments is 1200, and each of the last 10 payments is 900. Interest on the loan is at an annual effective rate of ii, i>0i>0. The interest portion of the 11th payment is twice the interest portion of the 21st payment. Calculate the interest portion of the 21st payment.

The interest portion of the 21st payment = 300

Explanation:

Interest component of 11th payment = Interest on the remaining loan at the end pf 10thperiod

Int(11) = 900(1 - v^20) + 300 (1 - v^10)

Int(11) = 1200 - 300v^10 - 900v^20

Interest compound of 21st payment = Interest on the remaining loan at the end of 20th period

Int(21) = 900 ( 1 - v^10)

Hence, Int(11) = 2 (Int(21)

1200 - 300v^10 - 900v^20 = 2 [900 ( 1 - v^10)]

1200 - 300v^10 - 900v^20 - 1800 + 1800v^10 = 0

-600 + 1500v^10 - 900v^20 = 0

9v20 - 15v^10 + 6 = 0

By solving above quadratic equation, we get

V^10 = 2/3

Int(21) = 900 ( 1 - V^10)

Int(21) = 900 ( 1 - 2/3)

Int(21) = 900 (1/3)

Int(21) = 300

odd consecutive integers are odd integers that follow each other. they have a difference of 2 between every two numbers. if n is an odd integer, then n, n+2, n+4 and n+6 will be odd consecutive integers. the first number in the pattern is always the variable on its own or in this case, "n". examples.

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