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A 10-year loan of 120,000 is to be repaid with payments at the end of each month. Interest is at an annual effective rate of 6.00%. The first monthly payment is 800. Each additional payment will be k more than the previous month payment. Find k.

Sagot :

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Answer:

k = 9.73.

Explanation:

So, from the question we are given the following parameters or information which are going to be useful in solving this problem, They are:

=> The total amount that was borrowed = 120,000, the number of years to repay the loan = 10 years, the annual interest rate = 6% and the first payment = 800.  

STEP ONE: Determine the present value.

We are going to need to determine the value for the compound interest which is given as: [(1 + 6%)^(1/12) - 1) × 12 = 5.841%

Therefore, the present value calculated as thus:

= 800/ (5.841 / 12) × ( 1 - 1/ (1 + 5.841/12)^(12 ×10)) = 72579.33.

STEP TWO: Determine the value of k.

The other value,r = k / (5.841 /12 × (1 + 5.841 /12)^(12 × 10))× [ (1 + 5.841 / 12)^(12 × 10 )- 1)/( 5.841 /12) -12 × 10).

= 4872.45 × k.

Therefore, 120,000 = 72579.33 + 4872.45 × k.

k = 9.73

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