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!
Best Answer
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 $97$.