# Amortization

This is an example of how sequences and series are used in “real life.”  It could be used in a calculus class or an advanced math class.

When you borrow money to buy a house or a car you are making a kind of loan called a mortgage. Paying off the loan is called amortizing the loan. The amount you borrow is called the principal. You agree to pay a fixed payment at regular intervals, usually monthly. The payment includes interest on the amount owed during the previous month plus a bit more to decrease the principal. Each month your payment decreases the remaining principal, so in the next month a little less goes to interest and a little more goes to paying off the principal. Let’s say you borrow A dollars (the principal) and agree to pay a monthly interest rate of r%, for a period of n months.

How much is the fixed (constant) monthly payment, P, so that after n months the loan and interest will be paid off?

After one month you make a payment and you now owe ${{A}_{1}}$ = the original amount plus the interest on this amount minus the payment, P. ${{A}_{1}}=A+Ar-P=A\left( 1+r \right)-P$

After the second payment is made you owe ${{A}_{2}}$ = ${{A}_{1}}$, plus the interest ${{A}_{1}r}$, minus the payment, P. ${{A}_{2}}=\left( A\left( 1+r \right)-P \right)+\left( A\left( 1+r \right)-P \right)r-P=A{{\left( 1+r \right)}^{2}}-P\left( 1+r \right)-P$

And likewise after the third month: ${{A}_{3}}=\left( A{{\left( 1+r \right)}^{2}}-P\left( 1+r \right)-P \right)+\left( A{{\left( 1+r \right)}^{2}}-P\left( 1+r \right)-P \right)r-P$ ${{A}_{3}}=A{{\left( 1+r \right)}^{3}}-P{{\left( 1+r \right)}^{2}}-P\left( 1+r \right)-P$

And after the fourth month a pattern emerges: ${{A}_{4}}=A{{\left( 1+r \right)}^{4}}-P{{\left( 1+r \right)}^{3}}-P{{\left( 1+r \right)}^{2}}-P\left( 1+r \right)-P$ ${{A}_{4}}=A{{\left( 1+r \right)}^{4}}-P\sum\limits_{k=0}^{3}{{{\left( 1+r \right)}^{k}}}$

And so on so that after n months the amount you owe is: ${{A}_{n}}=A{{\left( 1+r \right)}^{n}}-P\sum\limits_{k=0}^{n-1}{{{\left( 1+r \right)}^{k}}}$

The sigma expression represents a finite geometric series. The sum of such a series with a first term of 1 and a common ratio of R sum is given by $\displaystyle {{S}_{n}}=\frac{1-{{R}^{n}}}{1-R}$.

Therefore $\displaystyle {{A}_{n}}=A{{\left( 1+r \right)}^{n}}-P\left( \frac{1-{{\left( 1+r \right)}^{n}}}{1-\left( 1+r \right)} \right)=A{{\left( 1+r \right)}^{n}}-P\left( \frac{{{\left( 1+r \right)}^{n}}-1}{r} \right)$

But after n months you have paid off the loan and you owe An = 0 $\displaystyle 0=A{{\left( 1+r \right)}^{n}}-P\left( \frac{{{\left( 1+r \right)}^{n}}-1}{r} \right)$

This equation may be solved for P the monthly payment: $\displaystyle P=\frac{Ar{{\left( 1+r \right)}^{n}}}{{{\left( 1+r \right)}^{n}}-1}$.

The equation above is a formula for finding the payment on a mortgage. For example, to borrow $25,000 for a car at 3% per year (r = 0.0025 per month) for 5 years (n = 60) the monthly payment is $\displaystyle \frac{25,000\left( 0.0025 \right){{\left( 1+0.0025 \right)}^{60}}}{{{\left( 1+0.0025 \right)}^{60}}-1}=\449.22$. Of course, in “real life” someone looks up the amount on the internet and we all believe them. But now you can do it yourself and have some math fun at the same time. Advertisement ## 4 thoughts on “Amortization” 1. Pingback: Sequences | Teaching Calculus 2. Pingback: February 2022 | Teaching Calculus 3. Mr. Bill on said: Strangely, loans like these are typically referred to as “simple interest loans”. One explanation that I have seen is that computing the monthly payment is a “relatively simple process” which seems a bit presumptuous. More importantly, though, is that the monthly payments are such that their accumulated value will match what would be the ending balance of an interest-bearing account promising the same annual rate had the principal been invested and allowed to accrue interest compounded monthly over the life of the loan. Like 4. Jim on said: The interest rate of 3% annual interest is what is called a “nominal interest rate”. The precise monthly rate corresponding to 3% per year is (1.03)^(1/12) – 1 = .002466. A monthly rate of .0025 corresponds to a true annual rate of 3.042%.. At a true annual rate of 3% the monthly payment is$448.77.

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