# My Favorite One-Liners: Part 28

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them. Today’s quip is one that I’ll use when simple techniques get used in a complicated way.

Consider the solution of the linear recurrence relation

$Q_n = Q_{n-1} + 2 Q_{n-2}$,

where $F_0 = 1$ and $F_1 = 1$. With no modesty, I call this one the Quintanilla sequence when I teach my students — the forgotten little brother of the Fibonacci sequence.

To find the solution of this linear recurrence relation, the standard technique — which is a pretty long procedure — is to first solve the characteristic equation, from $Q_n - Q_{n-1} - 2 Q_{n-2} = 0$, we obtain the characteristic equation

$r^2 - r - 2 = 0$

This can be solved by any standard technique at a student’s disposal. If necessary, the quadratic equation can be used. However, for this one, the left-hand side simply factors:

$(r-2)(r+1) = 0$

$r=2 \qquad \hbox{or} \qquad r = -1$

(Indeed, I “developed” the Quintanilla equation on purpose, for pedagogical reasons, because its characteristic equation has two fairly simple roots — unlike the characteristic equation for the Fibonacci sequence.)

From these two roots, we can write down the general solution for the linear recurrence relation:

$Q_n = \alpha_1 \times 2^n + \alpha_2 \times (-1)^n$,

where $\alpha_1$ and $\alpha_2$ are constants to be determined. To find these constants, we plug in $n =0$:

$Q_0 = \alpha_1 \times 2^0 + \alpha_2 \times (-1)^0$.

To find these constants, we plug in $n =0$:

$Q_0 = \alpha_1 \times 2^0 + \alpha_2 \times (-1)^0$.

We then plug in $n =1$:

$Q_1 = \alpha_1 \times 2^1 + \alpha_2 \times (-1)^1$.

Using the initial conditions gives

$1 = \alpha_1 + \alpha_2$

$1 = 2 \alpha_1 - \alpha_2$

This is a system of two equations in two unknowns, which can then be solved using any standard technique at the student’s disposal. Students should quickly find that $\alpha_1 = 2/3$ and $\alpha_2 = 1/3$, so that

$Q_n = \displaystyle \frac{2}{3} \times 2^n + \frac{1}{3} \times (-1)^n = \frac{2^{n+1} + (-1)^n}{3}$,

Although this is a long procedure, the key steps are actually first taught in Algebra I: solving a quadratic equation and solving a system of two linear equations in two unknowns. So here’s my one-liner to describe this procedure:

This is just an algebra problem on steroids.

Yes, it’s only high school algebra, but used in a creative way that isn’t ordinarily taught when students first learn algebra.

I’ll use this “on steroids” line in any class when a simple technique is used in an unusual — and usually laborious — way to solve a new problem at the post-secondary level.

# Lychrel Numbers

A friend of mine posted the following on Facebook (with names redacted):

So [my daughter] comes home with this assignment:

For each number from 10 – 99, carry out the following process.

1.  If the number is a palindrome (e.g., 77), stop.
2.  Else reverse the number and add that to the original. E.g.: 45+54 = 99.
3.  If the result is not a palindrome, repeat step (2) with the result.
4.  Record the final palindromic result and the number of steps taken.

Most are simple.

• 56 + 65 = 110
• 110 + 011 = 121
• Stop. 2 steps taken.

The numbers 89 and 98 were given for extra credit, and they mysteriously explode, taking 24 steps. It made [my daughter] cry.

She wanted me to check her work, so I decided it was a good time to teach the wonders of Python, and we very quickly had a couple of simple functions to do the trick.

Well, you saw where this was going. How many steps does 887 take?

We’re up to 104000 steps so far, and Python is crying.

True or false: For a given n, the above algorithm completes in finite time?

I guess I’ve been living under a rock for the past 20 years, because I had never heard of this problem before. It turns out that numbers not known to lead to a palindrome are called Lychrel numbers. However, no number in base-10 has been proven to be a Lychrel number. The first few candidate Lychrel numbers (i.e., numbers that have not been proven to not be Lychrel numbers) are 196, 295, 394, 493, 592, 689, 691, 788, 790, 879, 887, 978, 986, 1495, 1497, 1585, 1587, 1675, 1677, 1765, 1767, 1855, 1857, 1945, 1947, 1997, 2494, 2496, 2584, 2586, 2674, 2676, 2764, 2766, 2854, 2856, 2944, 2946, 2996, 3493, 3495, 3583, 3585, 3673, 3675…

The above algorithm is called the 196-algorithm, after the smallest suspected Lychrel number.

For further reading, I suggest the following links and the references therein:

http://mathworld.wolfram.com/196-Algorithm.html

http://mathworld.wolfram.com/LychrelNumber.html

http://www.p196.org/

http://www.mathpages.com/home/kmath004/kmath004.htm (which contains a proof that 10110 is a Lychrel number in binary and that Lychrel numbers always exist in base $2^k$)

http://en.wikipedia.org/wiki/Lychrel_number

# Exponential growth and decay (Part 9): Amortization tables

This post is inspired by one of the questions that I pose to our future high school math teachers during our Friday question-and-answer sessions. In these sessions, I play the role of a middle- or high-school student who’s asking a tricky question to his math teacher. Here’s the question:

A student asks, “My father bought a house for $200,000 at 12% interest. He told me that by the time he fi nishes paying for the house, it will have cost him more than$500,000. How is that possible? 12% of $200,000 is only$24,000.”

Without fail, these future teachers don’t have a good response to this question. Indeed, my experience is that most young adults (including college students) have never used an amortization table, which is the subject of today’s post.

In the past few posts, we have considered the solution of the following recurrence relation, which is often used to model the payment of a mortgage or of credit-card debt:

$A_{n+1} = r A_n - k$

With this difference equation, the rate at which the principal is reduced can be simply computed using Microsoft Excel. This tool is called an amortization schedule or an amortization table; see E-How for the instructions of how to build one. Here’s a sample Excel spreadsheet that I’ll be illustrating below: Amortization schedule. My personal experience is that many math majors have never seen such a spreadsheet, even though they are familiar with compound interest problems and certainly have the mathematical tools to understand this spreadsheet.

Here’s a screen capture from the spreadsheet:

The terms of the loan are typed into Cells B1 (length of loan, in years), B2 (annual percentage rate), and B3 (initial principal). Cell B4 is computed from this information using the Microsoft Excel command $\hbox{PMT}$:

$=\hbox{PMT}(\hbox{B2}/12,\hbox{B1}*12,-\hbox{B3})$

This is the amount that must be paid every month in order to pay off the loan in the prescribed number of years. Of course, there is a formula for this:

$M = \displaystyle \frac{Pr}{12 \displaystyle \left[1 - \left( 1 + \frac{r}{12} \right)^{-12t} \right]}$

I won’t go into the derivation of this formula here, as it’s a bit complicated. Notice that this formula does not include escrow, points, closing costs, etc. This is strictly the amount of money that’s needed to pay down the principal.

The table, beginning in Row 8 of the above picture, shows how quickly the principal will be paid off. In row 8, the interest that’s paid for that month is  computed by

$=\hbox{B8} * \ \hbox{B}\\hbox{2}/12$

Therefore, the amount of the monthly payment that actually goes toward paying down the principal will be

$= \\hbox{B}\\hbox{4} - \hbox{C8}$

Column E provides an opportunity to pay something extra each month; more on this later. So, after taking into account the payments in columns D and E, the amount remaining on the loan is recorded in Cell F8:

$= \hbox{B8} - \hbox{D8} - \hbox{E8}$

This amount is then copied into Cell B9, and then the pattern can be filled down.

The yellow graph shows how quickly the balance of the loan is paid off over the length of the loan. A picture is worth a thousand words: in the initial years of the loan, most of the payments are gobbled up by the interest, and so the principal is paid off slowly. Only in the latter years of the loan is the principal paid off quickly.

So, it stands to reason that any extra payments in the initial months and years of the loan can do wonders for paying off the loan quickly. For example, here’s a screenshot of what happens if an extra $200/month is paid only in the first 12 months of the loan: A definite bend in the graph is evident in the initial 12 months until the normal payment is resumed in month 13. As a result of those extra payments, the curve now intersects the horizontal axis around 340. In other words, 20 fewer months are required to pay off the loan. Stated another way, the extra payments in the first year cost an extra $\200 \times 12 = \2400$. However, in the long run, those payments saved about $\536.82 \times 20 \approx \10,700$! # Exponential growth and decay (Part 8): Paying off credit-card debt via recurrence relations The following problem in differential equations has a very practical application for anyone who has either (1) taken out a loan to buy a house or a car or (2) is trying to pay off credit card debt. To my surprise, most math majors haven’t thought through the obvious applications of exponential functions as a means of engaging their future students, even though it is directly pertinent to their lives (both the students’ and the teachers’). You have a balance of$2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or$600 per year), how long will it take for the balance to be paid?

In previous posts, I approached this problem using differential equations. There’s another way to approach this problem that avoids using calculus that, hypothetically, is within the grasp of talented Precalculus students. Instead of treating this problem as a differential equation, we instead treat it as a first-order difference equation (also called a recurrence relation):

$A_{n+1} = r A_n - k$

The idea is that the amount owed is multiplied by a factor $r$ (which is greater than 1), and from this product the amount paid is deducted. With this approach — and unlike the approach using calculus — the payment period would be each month and not per year. Therefore, we can write

$A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$

Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months. The solution of this difference equation is

$A_n = r^n P - k \left( \displaystyle \frac{1 - r^n}{1-r} \right)$

A great advantage of using a difference equation to solve this problem is that the solution can be easily checked with a simple spreadsheet. (Indeed, pedagogically, I would recommend showing a spreadsheet like this before doing any of the calculations of the previous few posts, so that students can begin to wrap their heads around the notion of a difference equation before the solution is presented.)

To start the spreadsheet, I wrote “Step” in Cell A1 and “Amount” in Cell B1. Then I entered the initial conditions: $0$ in Cell A2 and $2000$ in Cell B2. (In the screenshot below, I changed the format of column B to show dollars and cents.) Next, I entered $=\hbox{A2}+1$ in Cell A3 and

$=\hbox{B2}*(1+0.25/12)-50$

in Cell B3. Finally, I copied the pattern in Cells A3 and B3 downward. Here’s the result:

After the formula the algebraic solution of the difference equation has been found, this can be added to the spreadsheet in a different column. For example, I added the header “Predicted Amount” in Cell D1. In Cell D2, I typed the formula

$=2000*\hbox{POWER}(1+0.25/12,\hbox{A2})-50*(1-\hbox{POWER}(1+0.25/12,\hbox{A2}))/(1-(1+0.25/12))$

Finally, I copied this pattern down the Column D. Here’s the result:

Invariably, when I perform a demonstration like this in class, I elicit a reaction of “Whoa…. it actually works!” Even for a class of math majors. Naturally, I tease them about this… they didn’t believe me when I used algebra, but now it has to be true because the computer says so.

Here’s the spreadsheet that I used to make the above pictures: CreditCardDebt.

# Exponential growth and decay (Part 7): Paying off credit-card debt via recurrence relations

The following problem in differential equations has a very practical application for anyone who has either (1) taken out a loan to buy a house or a car or (2) is trying to pay off credit card debt. To my surprise, most math majors haven’t thought through the obvious applications of exponential functions as a means of engaging their future students, even though it is directly pertinent to their lives (both the students’ and the teachers’).

You have a balance of $2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of$50 per month (or $600 per year), how long will it take for the balance to be paid? In previous posts, I approached this problem using differential equations. There’s another way to approach this problem that avoids using calculus that, hypothetically, is within the grasp of talented Precalculus students. Instead of treating this problem as a differential equation, we instead treat it as a first-order difference equation (also called a recurrence relation): $A_{n+1} = r A_n - k$ The idea is that the amount owed is multiplied by a factor $r$ (which is greater than 1), and from this product the amount paid is deducted. With this approach — and unlike the approach using calculus — the payment period would be each month and not per year. Therefore, we can write $A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$ Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months. A full treatment of the solution of difference equations belongs to a proper course in discrete mathematics. In the previous posts, I demonstrated how this difference equation could be solved by directly finding $A_1, A_2, A_3, \dots$ and looking for a pattern. In this post, I’d like to present an alternative method for deriving the solution. I’ll let the reader decide for him/herself as to whether this technique is pedagogically superior to the previous method. We will attempt to find a solution of the form $A_n = a r^n + b$, where $a$ and $b$ are unknown constants.Why do we guess the solution to have this form? I won’t dive into the details, but this is entirely analogous to constructing the characteristic equation of a linear differential equation with constant coefficients as well as using the method of undetermined coefficients to find a particular solution to a inhomogeneous linear differential equation with constant coefficients. Substituting $n+1$ instead of $n$, we find that $A_{n+1} = a r^{n+1} + b$. So we plug both of these into the difference equation: $A_{n+1} = r A_n - k$ $a r^{n+1} + b = r \left( a r^n + b \right) - k$ $a r^{n+1} + b = a r^{n+1} + r b - k$ $b = r b - k$ $k = (r-1) b$ $\displaystyle \frac{k}{r-1} = b$ We also use the fact that $A_0 = P$: $A_0 = a r^0 + b$ $P = a + b$ $P - b = a$ $\displaystyle P - \frac{k}{r-1} = a$ Combining these, we obtain the solution of the difference equation: $A_n = \displaystyle \left( P - \frac{k}{r-1} \right) r^n +\frac{k}{r-1}$ Unsurprisingly, this matches the solution that was obtained in the previous two posts (though the terms have been rearranged). # Exponential growth and decay (Part 6): Paying off credit-card debt via recurrence relations The following problem in differential equations has a very practical application for anyone who has either (1) taken out a loan to buy a house or a car or (2) is trying to pay off credit card debt. To my surprise, most math majors haven’t thought through the obvious applications of exponential functions as a means of engaging their future students, even though it is directly pertinent to their lives (both the students’ and the teachers’). You have a balance of$2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or$600 per year), how long will it take for the balance to be paid?

In previous posts, I approached this problem using differential equations. There’s another way to approach this problem that avoids using calculus that, hypothetically, is within the grasp of talented Precalculus students. Instead of treating this problem as a differential equation, we instead treat it as a first-order difference equation (also called a recurrence relation):

$A_{n+1} = r A_n - k$

The idea is that the amount owed is multiplied by a factor $r$ (which is greater than 1), and from this product the amount paid is deducted. With this approach — and unlike the approach using calculus — the payment period would be each month and not per year. Therefore, we can write

$A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$

Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months.

In yesterday’s post, I demonstrated that the solution of this recurrence relation is

$A_n = r^n P - k \displaystyle \left( \frac{1 - r^n}{1-r} \right)$.

Let’s now study when the credit card debt will actually reach $0. To do this, we see $A_n = 0$ and solve for $n$: $0 = r^n P - k \displaystyle \left( \frac{1 - r^n}{1-r} \right)$ $0 = r^n \left(P + \displaystyle \frac{k}{1-r} \right) - \displaystyle \frac{k}{1-r}$ $0 = r^n \left( P[1-r] + k \right) - k$ $k = r^n \left( P[1-r] + k \right)$ $\displaystyle \frac{k}{P[1-r] + k} = r^n$ $\displaystyle \ln \left( \frac{k}{P[1-r]+k} \right) = n \ln r$ $\displaystyle \frac{ \displaystyle \ln \left( \frac{k}{P[1-r]+k} \right) }{ \ln r} = n$ That’s certainly a mouthful. However, this calculation should be accessible to a talented student in Precalculus. Let’s try it out for $k = 50$, $P = 2000$, and $r = 1 + \displaystyle \frac{0.25}{12}$: Remembering that each compounding period is one month long, this corresponds to $86.897/12 \approx 7.24$ years, which is nearly equal to the value of $4\ln 6 \approx 7.17$ years when we solved this problem using differential equations under the assumption of continuous compound interest (as opposed to interest that’s compoounded monthly). # Exponential growth and decay (Part 5): Paying off credit-card debt via recurrence relations The following problem in differential equations has a very practical application for anyone who has either (1) taken out a loan to buy a house or a car or (2) is trying to pay off credit card debt. To my surprise, most math majors haven’t thought through the obvious applications of exponential functions as a means of engaging their future students, even though it is directly pertinent to their lives (both the students’ and the teachers’). You have a balance of$2,000 on your credit card. Interest is compounded continuously with a rate of growth of 25% per year. If you pay the minimum amount of $50 per month (or$600 per year), how long will it take for the balance to be paid?

In previous posts, I approached this problem using differential equations. There’s another way to approach this problem that avoids using calculus that, hypothetically, is within the grasp of talented Precalculus students. Instead of treating this problem as a differential equation, we instead treat it as a first-order difference equation (also called a recurrence relation):

$A_{n+1} = r A_n - k$

The idea is that the amount owed is multiplied by a factor $r$ (which is greater than 1), and from this product the amount paid is deducted. With this approach — and unlike the approach using calculus — the payment period would be each month and not per year. Therefore, we can write

$A_{n+1} = \displaystyle \left( 1 + \frac{0.25}{12} \right) A_n - 50$

Notice that the meaning of the 25% has changed somewhat… it’s no longer the relative rate of growth, as the 25% has been equally divided for the 12 months.

A full treatment of the solution of difference equations belongs to a proper course in discrete mathematics. However, this particular difference equation can be solved in a straightforward fashion that should be accessible to talented Precalculus students. Let’s use the above recurrence relation to try to find a pattern. For $n = 1$, we find

$A_1 = r A_0 - k = r P - k$.

For $n = 2$, we find

$A_2 = r A_1 - k$

$A_2 = r (rP - k) - k$

$A_2 = r^2P - rk - k$

$A_2 = r^2 P - k (1 + r)$

For $n = 3$, we find

$A_3 = r A_2 - k$

$A_3 = r \left[ r^2 P - k(1+r) \right] - k$

$A_3 = r^3 - rk(1+r) - k$

$A_3 = r^3 P - rk - r^2k - k$

$A_3 = r^2 P - k \left( 1 + r+r^2 \right)$

At this point, we can probably guess a pattern:

$A_n = r^n P - k \left( 1 + r + r^2 + \dots + r^{n-1} \right)$

Using the formula for a finite geometric series, this simplifies as

$A_n = r^n P - k \left( \displaystyle \frac{1 - r^n}{1-r} \right)$.

Indeed, though I won’t do it here, this can be formally proven using mathematical induction.

# Engaging students: Arithmetic sequences

In my capstone class for future secondary math teachers, I ask my students to come up with ideas for engaging their students with different topics in the secondary mathematics curriculum. In other words, the point of the assignment was not to devise a full-blown lesson plan on this topic. Instead, I asked my students to think about three different ways of getting their students interested in the topic in the first place.

I plan to share some of the best of these ideas on this blog (after asking my students’ permission, of course).

This student submission comes from my former student Erick Cordero. His topic, from Precalculus: arithmetic sequences.

“What interesting word problems using this topic can your students do now?”

There are many word problems we can do with arithmetic sequences but I am going to give one example that I believe students will understand. For this example, lets suppose that John Q, a pre-calculus student, has just bought a new phone from apple, but because of this new upgrade, Q’s parents are concern with the sum of money they will be paying for his monthly bill. Q’s first bill happens to be $65, his total after the second bill is$130, after the third bill the running sum is $195, if this pattern continues, how many months will it take for the total to reach$780? To solve this problem we would write the terms in a sequence starting with the first term being \$65 and up to three more terms. After writing out a few terms, I would expect the students to find the common difference between the terms and then compute the slope of the terms (I say slope because I hope they can see that this pattern is linear and therefore we can model the data using a linear equation and not just use the formula for arithmetic sequence but rather derive one ourselves). Then just like the students did in algebra one, they can use the point slope formula to come up with an equation for the sequence. I would explain to the students that now that we have the formula we can easily find the nth term that contains our sum, and this parallels the same process as having an x value and finding a corresponding y value and by using this process I can assure the students that the methods they learned in algebra are still important in pre-calculus.

“How can this topic be used in your students’ future courses in mathematics?”

Sequences and equations is a very important topic in mathematics, and unfortunately many students that take pre-calculus in high school will never get to experience how sequences evolve from simple arithmetic sequences to the more powerful ones in calculus II. Sequences are often overlook by students in pre-calculus (high school) because it is different from what they have encountered in their math career thus far, but maybe if we show students how this topic evolves in calculus II then they will pay more attention to it (Or they will forget it more since many students will not take calculus II). But from an educators’ standpoint, we understand how important sequences are. In calculus II teachers teach students how the elementary ideas they learned in pre-calculus are now used in calculus applications. One of these ideas is called a power series. Power series are fundamental to the study of calculus because they provide a way to represent some of the most important functions in our field. Power series are also useful in physics and chemistry. We also have Taylor Series, which have been regarded by some as the most interesting topic in calculus II. It is here, in calculus II where we see the true power of sequences and for some of us, that random topic in pre-calculus about sequences starts to make sense. Sequences is a topic that in rooted deep in the heart of mathematics and we should tell our students in pre-Cal, or algebra, how important this topic is as they go deeper into their math or science careers.

“How can technology be used to effectively engage students with this topic?”

One website that I have often visit is Khan Academy, and I would encourage my students to do the same. I like this website because unlike some of the YouTube videos, these videos are more engaging and interesting. The person doing the videos is also more professional and has an understanding of mathematics beyond some of the YouTube clips I have encountered. The quality of this website is the best I have seen. I also like how Sal Khan (the person doing the videos) uses a lot of colors in his videos because it helps the students distinguish information. This is another reason why YouTube is sometimes not a great idea. Some of the videos are of people solving math problems on a white board, if that’s the point then why show the clip in the first place? Students do not want to see that, I will do enough of that. I have said enough bad things about YouTube, and hence it is only fair that I now show something positive from it.

The above is a YouTube clip from Khan Academy where Khan does a problem trying to find the 100th term of a sequence. Khan Academy is great place were students can see more examples of certain classroom topics but of course this is not something to replace classroom work but rather another option to engage students with.

# Pedagogical thoughts about sequences and series (Part 2)

After yesterday’s post about arithmetic and geometric sequences, I’d like to contribute some thoughts about teaching this topic, based on my own experience over the years.

1. Some students really resist the subscript notation $a_n$ when encountering it for the first time. To allay these concerns, I usually ask my students, “Why can’t we just label the terms in the sequence as $a$, $b$, $c$, and so on?” They usually can answer: what if there are more than 26 terms? That’s the right answer, and so the $a_n$ is used so that we’re not limited to just the letters of the English alphabet.

Another way of selling the $a_n$ notation to students is by telling them that it’s completely analogous to the $f(x)$ notation used more commonly in Algebra II and Precalculus. For a “regular” function $f(x)$, the number $x$ is chosen from the domain of real numbers. For a sequence $a_n$, the number $n$ is chosen from the domain of positive (or nonnegative) integers.

2. The formulas in Part 1 of this series (pardon the pun) only apply to arithmetic and geometric sequences, respectively. In other words, if the sequence is neither arithmetic nor geometric, then the above formulas should not be used.

While this is easy to state, my observation is that some students panic a bit when working with sequences and tend to use these formulas on homework and test questions even when the sequence is specified to be something else besides these two types of sequences. For example, consider the following problem:

Find the 10th term of the sequence $1, 4, 9, 16, \dots$

I’ve known pretty bright students who immediately saw that the first term was $1$ and the difference between the first and second terms was $3$, and so they answered that the tenth term is $1 + (10-1)\times 3 = 28$… even though the sequence was never claimed to be arithmetic.

I’m guessing that these arithmetic and geometric sequences are emphasized so much in class that some students are conditioned to expect that every series is either arithmetic or geometric, forgetting (especially on tests) that there are sequences other than these two.

3. Regarding arithmetic sequences, sometimes it helps by giving students a visual picture by explicitly make the connection between the terms of an arithmetic sequence and the points of a line. For example, consider the arithmetic sequence which begins

$13, 16, 19, 22, \dots$

The first term is $13$, the second term is $16$, and so on. Now imagine plotting the points $(1,13)$, $(2,16)$, $(3,19)$, and $(4,22)$ on the coordinate plane. Clearly the points lie on a straight line. This is not surprising since there’s a common difference between terms. Moreover, the slope of the line is $3$. This matches the common difference of the arithmetic sequence.

4. In ordinary English, the words sequence and series are virtually synonymous. For example, if someone says either, “a sequence of unusual events” or “a series of unusual events,” the speaker means pretty much the same thing

However, in mathematics, the words sequence and series have different meanings. In mathematics, an example of an arithmetic sequence are the terms

$1, 3, 5, 7, 9, \dots, 99$

However, an example of an arithmetic series would be

$1 + 3 + 5 + 7 + 9 + \dots + 99$

In other words, a sequence provides the individual terms, while a series is a sum of the terms.

When teaching this topic, I make sure to take a minute to emphasize that the words sequence and series will mean something different in my class, even though they basically mean the same thing in ordinary English.