Introduction

Welcome to unit 2! In this topic, you’ll learn how to represent, analyze, and compare two fundamental types of sequences: arithmetic and geometric. These sequences appear frequently in both mathematical problems and real-world contexts, such as financial planning, population modeling, and patterns of growth and decay.

What is a sequence?

A sequence is a list of numbers whose domain is all non-negative integers and whose values are real numbers. The notation $a_{n}$ denotes the $n$ -th term of the sequence. Because the input $n$ must be a non-negative integer, a sequence is discrete, meaning it only exists at specific, distinct values, which also means that nothing exists between those values/points. Typically, on a graph, functions appear as one connected and continuous line; however, if it is discrete, the graph will appear as individual points with gaps between them.

For example, if $a (n) = 2 n + 5$ for $n = 0, 1, 2, \dots,$ we plot ONLY the points $(0, 5)$ , $(1, 7)$ , $(2, 9), ....$ . We can think about sequences like they are functions; we can use terms like domain or formulas just as with other functions. In addition, it is recommended to stick to whole numbers as it will make calculations much more easier and not have to do any unnecessary work, However, you must keep in mind the main differences, which are that the domain is restricted to exclusively the positive integers for sequences and that the sequence is discrete and thus the graph consists only of separate points rather than a continuous line. So, in other words, think of sequences like functions except they don’t connect together; they are simply individual points.

Arithmetic Sequences

An arithmetic sequence is one of the two sequences you will learn in AP Precalculus; it is a sequence in which successive terms differ by a constant amount. In other words, there is a fixed “d” value (the common difference) such that each term is the previous term plus d: Formally, $a_{n + 1} = a_{n} + d$ where $n$ is an non-negative integer ( $\geq 0$ ) and $d$ is a real number.

The CollegeBoard assumes the first term is $a_{0}$ ; some other websites may use $a_{1}$ as their starting term, but for this class, you will use $a_{0}$ as your first term. This style is called zero-based indexing.

We can view arithmetic sequences as a linear function, but one where $n$ (or the domain) is only defined on all the non-negative integers.

In order to find the common difference $d$ (given an arithmetic sequence), we can use the fact that each consecutive term is separated by that difference $d$ . For example, the sequence

$5, 8, 11, 14 \dots$

is an arithmetic sequence with a common difference $d = 3$ , because each consecutive term increases by a difference of 3. Listing the terms shows the pattern clearly.

$a_{0} = 5$

$a_{1} = a_{0} + 3 = 5 + 3 = 8$

$a_{2} = a_{1} + 3 = 8 + 3 = 11$

Here, you can see that each time the term number $n$ increases by one, the value gets $1 \cdot 3$ added.

We can also write $a_{2}$ in terms of $a_{0}$ instead of $a_{1}$ : $a_{2} = a_{0} + 2 \cdot 3 = 5 + 2 \cdot 3 = 11$ .

Following this pattern, we can get the general form for an arithmetic sequence: $a_{n} = a_{0} + n \cdot d$ . Equivalently, if we know the value at some index $k$ (say $a_{k}$ ), and want the value at index $n$ , we add $d n - k$ times: $a_{n} = a_{k} + d (n - k)$

It is important of us to learn and practice what we have just learnt, so let us work through some practice problems!

$a_{n + 1} + d, n = 0, 1, 2, \dots$

EXAMPLE 1:

Question: The following numbers are the first four terms of an arithmetic sequence

$4, 1.5, - 1, - 3.5$

Find the arithmetic sequence that best follows the given four terms.

Take some time to solve this by yourself, and then we will work together.

Firstly, remember that the general form of an arithmetic sequence is $a_{n} = a_{0} + n d$ . The first term ( $a_{0}$ ) would be $4$ , so $a_{0} = 4$ . As we continue down the series, notice that each next term is $2.5$ less than the one before it. $1.5 - 4 = - 2.5$ , $- 1 - 1.5 = - 2.5$ . As such, the common difference is $d = - 2.5$ .

Plug these values into the general form to get $a_{n} = 4 + (- 2.5) n = 4 - 2.5 n$

EXAMPLE 2:

Find the explicit form for the arithmetic sequence $a$ if $a_{3} = 94$ and $a_{6} = 85$

Again, take some time to figure this problem out.

Recognize that we have another formula that can be used for specifically the explicit form of a sequence, namely $a_{n} = a_{k} + d (n - k)$ . This time, we are given $a_{6}$ and $a_{3}$ . In this case, we can say that $n = 6$ and $k = 3$ so that $a_{n} = 85$ and $a_{k} = 94$ .

Since the only information we do not have is the common difference $d$ , that will be the final constant to solve for. Plug these values into the equation to get $85 = 94 + d (6 - 3) \to 85 = 94 + 3 d$ . Afterwards, solve for $d$ to get that $d = - 3$ .

The sequence will be in terms of $n$ , so we will not plug in $a_{n} = 85$ or $n = 6$ . However, plugging in $a_{k} = 94$ , $k = 3$ , and $d = - 3$ will give the arithmetic sequence.

$a_{n} = 94 + (- 3) (n - 3) = 94 - 3 (n - 3)$

Don’t simplify or do anything to it, as the problem asked for the explicit form that we have provided above.

EXAMPLE 3:

Question: A savings account has an initial deposit of $100, and every month $20 is deposited into the account. Write the arithmetic sequence that best describes this scenario and then find the amount of money in the account after six months.

Once again, take a little time to work this out in your head.

From here, we see that at the zeroth month ( $n = 0), t h e a cco u n t ha s$ 100. This means that in our sequence, $a_{0} = 100$ . Next, the balance is changing by $20 each month. This change is our common difference.

To write an arithmetic sequence, we will use the general form $a_{n} = a_{0} + n d$ . $a_{0} = 100$ and $d = 20$ (the common difference). Plugging this in gives us $a_{n} = 100 + 20 n$ .

To find the amount of money after six months, simply plug in $n = 6$ into our formula to get that $a_{6} = 100 + 20 (6) = 100 + 120 = 220$ . As such, the account will have $220 after it opens.

Geometric Sequences

A geometric sequence is a sequence in which successive terms differ multiplicatively by a constant factor (known as the common ratio). That is, there is a fixed number $r$ such that

$g_{n + 1} = r^{g_{n}}, n = 0, 1, 2, ...$

Each term is the previous term multiplied by $r$ . Equivalently, each term divided by its previous term is constant. If $r > 0$ , then $\frac{g _{n}}{g _{n - 1}} = r$ for all $n$ . For example, the sequence

$3, 6, 12, 24 \dots$

is a geometric sequence with a common ratio r = 2, since each term is twice the value of the previous term, which can be more clearly seen in below:

$3 \cdot 2^{0} = 3, 3 \cdot 2^{1} = 6, 3 \cdot 2^{2} = 12, 3 \cdot 2^{3} = 24 \dots$

Here, you can see that each time the term number $n$ increases by one, the value is doubled, or multiplied by two. This multiplicative effect is a result of our common ratio $r$ , which represents whatever number is being multiplied to the previous term of a geometric sequence.

Quick note: $g_{n}$ is the output value that we are finding, which can be similar to the $y$ -value of a function.

Writing out terms:

$g_{0} = 3, g_{1} = 3 \cdot 2 = 6, g_{2} = 3 \cdot 2^{2} = 12, g_{3} = 3 \cdot 2^{3} = 24 \dots$

Shows the pattern. In general, starting from $g_{0}$ and multiplying $r$ an $n$ number of times gives

$g_{n} = g_{0} \cdotr^{n}$

More generally, if we know $g_{k}$ and want $g_{n}$ , we multiply by $r$ a total of $n - k$ times:

$g_{n} = g_{k} \cdotr^{n - k}$

This formula follows because each step multiplies the previous term by $r$ , so $g_{1} = g_{0} \cdotr, g_{2} = g_{0} \cdotr^{2}$

EXAMPLE 1:

A geometric sequence has a first term of $4$ and a common ratio of $3$ . What is the fourth term?

Take a moment to work it out on paper.

Since we have the first term and the common ratio, we can use the formula $g_{n} = g_{0} \cdotr^{n}$ . Since $4$ is the first term of the sequence, $g_{0} = 4$ . The common ratio is $3$ , so $r = 3$ . Plug these in to get the sequence, $g_{n} = 4 \cdot 3^{n}$ .

To find the fourth term, we can simply plug in $n = 3$ into the sequence. The reason why we do not plug in $n = 4$ is because the first term is represented by $n = 0$ . First being zero, then fourth being three makes sense. It is okay to plug in $n = 4$ in the circumstance that you set $g_{1} = 4$ instead.

$g_{3} = 4 \cdot 3^{3} = 4 \cdot 27 = 108$ .

EXAMPLE 2:

In a geometric sequence, it is given that $g_{2} = 5$ and $g_{n} = - 2 r^{n - 1}$ . Find the common ratio and write the formula for the $n$ th term of the sequence.

Take a little more time to solve this and figure it out.

First of all, we can start this problem by substituting our given value of $g_{2} = 5$ into our geometric sequence formula. This gets us $5 = - 2 \cdot r^{2 - 1}$ , which can simplify to $5 = - 2 \cdot r^{1}$ .

Now let’s find $r$ !

At this stage, it’s pretty much just algebra, so we got this!

$5 = - 2 r$

$- \frac{5}{2} = r$

$r = - \frac{5}{2}$

This is the value of $r$ , the ratio of our geometric sequence. The next part of the problem states to write the formula for the $n$ th term, which is simply incorporating our newly found $r$ into the given equation above.

$g_{n} = - 2 (- \frac{5}{2})^{n - 1}$

That’s our answer, nice work!

Conclusion

In arithmetic sequences, each term increases by the same fixed amount. No matter how far along the sequence you go, the difference between consecutive terms remains constant. Geometric sequences multiply and grow terms instead of increasing by the same fixed amount. This means that as the sequence progresses, the amount by which each term increases becomes larger and larger in a multiplicative way. In other words, arithmetic growth adds the same amount each time, while geometric growth multiplies by the same factor with each successive step.

Let’s check comprehension. Problems are listed in increasing difficulty:

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