Tag Archives: amortization

Sequences

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Here is a list of past posts on the topics of sequences and series that I hope you find interesting and useful. The first two are suitable for precalculus students.

The first uses sequences and series for a very practical aim that affects almost everyone sometime in their life: paying off a loan. The next gives a good, and I hope, understandable explanation of what an irrational number is.

Amortization. When you have a mortgage on your home or your car, you make the same payment every month. Part of the money pays the interest on the outstanding balance for the last month; the rest pays down the principal so there is less to pay interest on next month. How is the payment computed?

A Lesson on Sequences. What is the square root of two? Really, what is it? This post is an outline of a lesson finding a sequence of numbers that converges to a specific number known in advance and by doing so defines that number.

The next three posts deal with convergences tests and are of interest to BC students at this time of year.

Reference Chart. An outline of the various convergence tests, and their hypotheses (when you can use them).

These two posts answer the question in their titles:

Which Convergence Test Should I Use? Part 1: Pretty much anyone you want!

Which Convergence Test Should I Use? Part 2: Specific hints and a discussion of the usefulness of absolute convergence

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Amortization

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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.amortization

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.