# Solving for the amount of level payments proportional to interest due from sample SOA exam

this is a problem from practice exam FM that I am studying for:

A 20-year loan of 1000 is repaid with payments at the end of each year. Each of the first ten payments equals 150% of the amount of interest due. Each of the last ten payments is X. The lender charges interest at an annual effective rate of 10%. Calculate X.

I've started this problem by finding the loan balance at time 10 after the first 10 interest payments using a brute force method for each year

end of year 1 payment $$=1000(0.10)(1.5)=150$$

amount at end of year 1 $$=1000(0.10)-150=950$$

year 2 payment $$=950(0.10)(1.5)=142.5$$...etc

I know this method would take too long on an actual exam. Is there an easier method to do this? I know once I find the loan amount at time 10 I can easily take use an annuity with level payments to finish calculating X, but this is the part that I am stuck at.

Thank you!

This is a good start. Let denote the amount of the loan at the end of year t as $$A_t$$. Then we have $$A_1=950=1000\cdot 0.95^1$$ and $$A_2=950-142.5+95=902.5$$ This is equal to $$1000\cdot 0.95^2$$. The pattern is clear now.

Up to the end of the $$\textrm{10th}$$ year we have $$A_{10}=1000\cdot 0.95^{10}$$. If we choose the end of the $$\textrm{20th}$$ year as $$\textrm{reference date}$$ we have to compound this value ten times: $$1000\cdot 0.95^{10}\cdot 1.1^{10}$$

Next we have to repay an amount of $$X$$ for the next 10 years. Here we use the geometric series. End of the $$\textrm{10th}$$ year is equal to the beginning of the $$\textrm{11th}$$ year. The (future) value at the reference date is $$\sum\limits_{k=0}^{^9}X\cdot 1.1^k=X\cdot \frac{1.1^{10}-1}{0.1}$$. This is the value of the second 10 payments at the end of the 20th year. Therefore the equation is

$$1000\cdot 0.95^{10}\cdot 1.1^{10}=X\cdot \frac{1.1^{10}-1}{0.1}$$

$$1000\cdot 0.95^{10}$$ can be seen as a present value of a loan and we have to repay 10 years an amount of $$X$$. Then

$$X= 1000\cdot 0.95^{10}\cdot 1.1^{10}\cdot \frac{0.1}{1.1^{10}-1}\approx 97.44$$ Rounded to the nearest integer we get $$97$$.

• Thanks for downvoting without leaving a comment. I don´t understand people with such a behaviour. Commented Apr 13, 2020 at 14:22