Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A loan for $50,000$ has level payments to be made at the end of each year for 10 years at an annual rate of $9 \%$. (a) Find the balance at the end of $3$ years. (b) Find the interest paid in the third payment.

So I think this involves using amortization. That is, one pays the interest due first and then the principal afterwards. In other words, one covers the "creeping costs" first, and then pays the full amount. So this is a immediate annuity. So the interest due would be $50000(1.09)^{10}$?

share|cite|improve this question
up vote 2 down vote accepted

Conceptually, the way amortization of loan payments works is that the lender should come out even at the end of the loan when you account for future value of money based on the loan's interest rate. If the lender loaned $A$ dollars at an interest rate per period $r$ over $n$ periods, the future value (at the end of the loan) of the money lent is $A(1+r)^n$. If the payment is $p$, then the future value of the first payment is $p(1+r)^{n-1}$ (paid at the end of the first period), for the second is $p(1+r)^{n-2}$, for the $k$th payment is $p(1+r)^{n-k}$, up to the last payment which is paid at the end of the loan so its future value is $p(1+r)^0=p$. If you take the sum of these future values of all the payments, it should be equal to the future value of the original loan. This equation can be solved for $p$ in terms of the general $A$, $r$, and $n$ to get a generalized amortization formula.

Now, on a period-to-period basis, the balance is defined by the initial balance $b_0=A$ and the recurrence relation $b_{k}=(1+r)b_{k-1}-p$. That is, the change from one period to the next is to add the interest due and subtract the payment. Knowing the parameters of the loan and the payment, you can use this to find the balance after a specific number of periods, as well as the breakdown of interest and principal in each payment.

share|cite|improve this answer

You need to determine the payment so you come out even at the end of 10 years. There should be a formula in your text for that. If I recall correctly, it looks like payment = interest * principal/(1-(1+i)^-t), but I don't know what symbols you use for the terms. Then you are right, as each payment is made you deduct the interest that has accrued in the past year, then apply the rest to the principal. The interest due in the third year will be somewhat less than 9% of 50000 because the principal will have been reduced by the first two year's payments.

share|cite|improve this answer

I am not sure if this is a homework. So I will give you the answers and you figure out yourself.

Let $i=0.09, v=(1+i)^{-1}=0.917431192$.

Answer for (a): $$50000\frac{1-v^7}{1-v^{10}}$$ Answer for (b): $$50000\times 0.09\times\frac{1-v^8}{1-v^{10}}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.