Arithmetic Sequences

Learning goals

Index a sequence's terms by position
Find the common difference from consecutive terms
Apply $a_n = a_1 + (n-1)d$ for any term
Write the recursive rule beside the explicit one
Recover $d$ from two terms with $a_k = a_m + (k-m)d$
Solve for $n$ to test whether a value appears

Naming the terms of a sequence

A sequence is an ordered list of numbers. Order is part of what a sequence is: the list $2, 5, 8, 11$ is a different sequence from $11, 8, 5, 2$ , even though the same four numbers appear. Each number in the list is called a term.

Because the order matters, we label the terms by their position. We write $a_1$ for the first term, $a_2$ for the second, $a_3$ for the third, and in general $a_n$ for the term in position $n$ . The small number written below the line is the subscript, and it is just the position counter, not a multiplication. For the sequence $2, 5, 8, 11$ ,

a_1 = 2, \qquad a_2 = 5, \qquad a_3 = 8, \qquad a_4 = 11.

Another way to see this: a sequence is a function whose inputs are the positions $1, 2, 3, \ldots$ and whose outputs are the terms. Instead of writing $f(1)$ , $f(2)$ , $f(3)$ the way we did for ordinary functions, we write $a_1$ , $a_2$ , $a_3$ . The subscript notation is only a compact way to write “the output at position $n$ .” The position $n$ is always a counting number ( $1, 2, 3, \ldots$ ), never a fraction or a negative, because there is no “term number one-half.”

What makes a sequence arithmetic

An arithmetic sequence is a sequence in which you get each term by adding the same fixed number to the term before it. That fixed number is called the common difference, written $d$ . It is the gap between one term and the next, and in an arithmetic sequence that gap never changes.

To find $d$ , subtract any term from the one that follows it:

d = a_2 - a_1 = a_3 - a_2 = a_4 - a_3 = \cdots

Take $3, 7, 11, 15, 19$ . The differences are $7 - 3 = 4$ , then $11 - 7 = 4$ , then $15 - 11 = 4$ , then $19 - 15 = 4$ . Every gap is $4$ , so this is arithmetic with $d = 4$ . The common difference can be negative, which makes the sequence decrease: $20, 17, 14, 11$ has $d = 17 - 20 = -3$ , since each term is three less than the one before. It can also be a fraction or a decimal. The common difference can even be zero: then every term is the same and the sequence is constant, which still counts as arithmetic (it just never changes).

To test whether a sequence is arithmetic, check that all the consecutive differences agree. The sequence $2, 4, 8, 16$ is not arithmetic: the first gap is $4 - 2 = 2$ but the second is $8 - 4 = 4$ , and a single mismatch is enough to disqualify it.

An arithmetic sequence marches along the number line in equal steps of size d. Reaching the 5th term takes 4 jumps from the first, which is the reason the nth term takes n-1 of them.

A formula for the nth term

Adding $d$ over and over does reach any term, but it is slow. To find the $100$ th term you would add $d$ ninety-nine times. There is a shortcut, and the picture above already shows why it works. To land on a later term you take a whole number of equal jumps from the first term, so you only need to count the jumps.

The $n$ th term is $a_n = a_1 + (n-1)d$ #

Start at the first term and add the common difference one step at a time. The second term is the first term plus one $d$ :

a_2 = a_1 + d.

The third term is the second plus another $d$ , so it is the first term plus two $d$ ‘s:

a_3 = a_2 + d = (a_1 + d) + d = a_1 + 2d.

The fourth term adds one more $d$ , which makes three $d$ ‘s in all:

a_4 = a_3 + d = a_1 + 3d.

A pattern is now plain. The number of $d$ ‘s added is always one less than the position of the term. The second term carries one $d$ , the third carries two, and the fourth carries three. The reason is easy to say in words. To travel from the first term to the $n$ th term you take $n - 1$ single steps, and every step adds exactly $d$ . That count is right because those steps land you on the second term, then the third, and so on up to the $n$ th. Adding $n - 1$ copies of $d$ to the first term gives

a_n = a_1 + (n-1)\,d.

This closed formula reaches any term straight from its position $n$ , with no need to build the terms before it.

The formula has a clean meaning. You begin at $a_1$ , and then you make $(n - 1)$ steps of size $d$ , so the term is the starting value plus the total of those steps, $(n-1)d$ . Everything about an arithmetic sequence is captured by just two numbers: the first term $a_1$ and the common difference $d$ .

Worked example 1 Find the 15th term

Find the $15$ th term of the sequence $4, 11, 18, 25, \ldots$

First read off the two numbers that define the sequence. The first term is $a_1 = 4$ . The common difference is the gap between consecutive terms, $11 - 4 = 7$ , so $d = 7$ .

Now use the formula with $n = 15$ . The number of steps is $n - 1 = 14$ :

a_{15} = a_1 + (15 - 1)d = 4 + 14 \times 7.

Work out the product before the sum, $14 \times 7 = 98$ , then add:

a_{15} = 4 + 98 = 102.

So the $15$ th term is $102$ . Notice you never had to list the first fourteen terms.

Check your understanding

An arithmetic sequence has first term $a_1 = 40$ and common difference $d = -3$ . What is the $10$ th term $a_{10}$ ?

13

10

16

7

The recursive rule

The formula $a_n = a_1 + (n-1)d$ is called the explicit (or closed) form, because it gives a term directly from its position. There is a second, equally natural way to describe the same sequence, called the recursive form. It states how each term is built from the one right before it:

a_n = a_{n-1} + d \qquad \text{for } n \ge 2,

together with a stated first term $a_1$ . Read aloud, it says exactly the definition of arithmetic: “each term is the previous term plus the common difference.” The recursive rule is often the most honest description of a real process (you deposit $25$ more dollars onto last month’s balance). But to reach a far-off term with the recursive rule you must climb through every term in between. The explicit formula is the tool for jumping ahead; the recursive rule is the tool for describing the step.

Climbing through the terms is worth doing once by hand, because it is what makes the $n - 1$ stop being an arbitrary detail. The figure below is a single step on a number line. Both halves of that step are under your control: the number the step starts from, and the number added to it.

Set the start to $-5$ and the number added to $3$ . The arrow lands on $-2$ , so a sequence with $a_1 = -5$ and $d = 3$ has $a_2 = -2$ . Now move the start to $-2$ , where you just landed, and take the same step again. You reach $1$ , which is $a_3$ . Two more re-starts, from $1$ and then from $4$ , carry you to $4$ and then to $7$ . Count the steps you took: four of them, to get from $a_1$ to $a_5$ . That is where $n - 1$ comes from. That count is off by one from the position for the plainest possible reason, which is that standing on the first term costs no steps at all. Then set the number added to $-3$ and walk the same route in reverse. Four steps of $-3$ return you from $7$ to $-5$ , which is why a negative common difference runs the sequence downward without changing anything else about the counting.

Number line walk

Start at Then add

-5 + 3 = -2. Adding a positive number moves 3 units to the right.

One step on a number line, with both the number it starts from and the number added to it under your control. Restarting the same step from wherever the last one landed is the recursive rule of an arithmetic sequence, taken one term at a time.

The same counting argument works between any two terms, not only from the first. To get from the $m$ th term to the $k$ th term you take $k - m$ steps of size $d$ , so

a_k = a_m + (k - m)\,d.

This version is what you reach for when you know two terms and want to recover the common difference.

Worked example 2 Find $d$ and $a_1$ from two terms

In an arithmetic sequence the third term is $a_3 = 20$ and the seventh term is $a_7 = 44$ . Find the common difference and the first term.

Between the third term and the seventh term there are $7 - 3 = 4$ steps, each of size $d$ . Using $a_k = a_m + (k - m)d$ ,

a_7 = a_3 + (7 - 3)d \quad\Longrightarrow\quad 44 = 20 + 4d.

Solve for $d$ by subtracting $20$ and dividing by $4$ :

4d = 24 \quad\Longrightarrow\quad d = 6.

Now back up from the third term to the first. The first term is two steps below the third, so $a_1 = a_3 - 2d$ :

a_1 = 20 - 2 \times 6 = 20 - 12 = 8.

The sequence is $8, 14, 20, 26, 32, 38, 44, \ldots$ , and a quick check confirms $a_3 = 20$ and $a_7 = 44$ .

Working backward: which term is it?

The explicit formula also answers the reverse question. Instead of asking “what is term number $n$ ,” you can ask “which term equals a given value,” and you find out by solving for $n$ . Because $n$ must be a counting number, this also tells you whether a value appears in the sequence at all.

Worked example 3 Which term equals a given value?

For the sequence $5, 8, 11, 14, \ldots$ , decide which term equals $98$ , and whether $100$ ever appears.

Here $a_1 = 5$ and $d = 3$ , so the explicit formula is $a_n = 5 + (n - 1)\,3$ . Set it equal to $98$ and solve for $n$ :

5 + 3(n - 1) = 98 \quad\Longrightarrow\quad 3(n - 1) = 93 \quad\Longrightarrow\quad n - 1 = 31.

So $n = 32$ , which is a whole number, and $98$ is the $32$ nd term.

Now try $100$ the same way:

5 + 3(n - 1) = 100 \quad\Longrightarrow\quad 3(n - 1) = 95 \quad\Longrightarrow\quad n - 1 = \tfrac{95}{3}.

Since $\tfrac{95}{3}$ is not a whole number, there is no position $n$ that lands on $100$ , so $100$ is not a term of this sequence. A value belongs to the sequence only when solving for $n$ gives a positive whole number.

A real-world pattern

Worked example 4 Seats in a theater

A theater has $18$ seats in its front row, and each row behind it holds $2$ more seats than the row in front. How many seats are in row $12$ , and which row is the first to hold $54$ seats?

The seat counts by row form an arithmetic sequence with first term $a_1 = 18$ (the front row) and common difference $d = 2$ . The number of seats in row $n$ is

a_n = 18 + (n - 1)\,2.

For row $12$ , put $n = 12$ , so there are $n - 1 = 11$ steps:

a_{12} = 18 + 11 \times 2 = 18 + 22 = 40 \text{ seats}.

To find the row with $54$ seats, set $a_n = 54$ and solve for $n$ :

18 + 2(n - 1) = 54 \quad\Longrightarrow\quad 2(n - 1) = 36 \quad\Longrightarrow\quad n - 1 = 18.

So $n = 19$ , and row $19$ is the one with $54$ seats. The same two ideas, plug in a position or solve for a position, handle almost every arithmetic-sequence question you will meet.

Common mistakes

Using $n$ instead of $n - 1$ steps. The $n$ th term adds the common difference $n - 1$ times, not $n$ times. The first term already sits at position $1$ with zero steps taken, so $a_n = a_1 + (n-1)d$ . Writing $a_1 + nd$ overshoots by one full step.
Subtracting the difference the wrong way. The common difference is a later term minus an earlier term, $d = a_2 - a_1$ . Reversing it flips the sign, turning a rising sequence into a falling one.
Losing the sign of a negative $d$ . When the sequence decreases, $d$ is negative, and $(n-1)d$ must stay negative through the whole computation. It helps to keep the negative in parentheses, as in $40 + 9 \times (-3)$ .
Assuming every value appears. Solving $a_n = V$ tells you the position only if $n$ comes out a positive whole number. If $n$ is a fraction, the value simply is not a term of the sequence.
Confusing the position with the term. In “the $12$ th term is $40$ ,” the number $12$ is the position $n$ and $40$ is the term $a_n$ . Feeding $40$ back in as if it were the position is a common slip in word problems.

Key takeaways

A sequence is an ordered list of terms $a_1, a_2, a_3, \ldots$ , where the subscript $n$ is the position; it is a function whose inputs are the counting numbers.
A sequence is arithmetic when consecutive terms differ by a constant common difference $d = a_2 - a_1 = a_3 - a_2 = \cdots$ , which may be positive, negative, zero, or fractional.
The explicit formula is $a_n = a_1 + (n-1)d$ : start at $a_1$ and take $n - 1$ steps of size $d$ .
The recursive rule is $a_n = a_{n-1} + d$ with $a_1$ given, describing one step at a time.
Between any two terms, $a_k = a_m + (k - m)d$ , which recovers $d$ from two known terms.
To find which term equals a value $V$ , solve $a_n = V$ for $n$ ; the value is in the sequence only if $n$ is a positive whole number.

Arithmetic Sequences

Learning goals

Naming the terms of a sequence

What makes a sequence arithmetic

A formula for the nth term

The recursive rule

Working backward: which term is it?

A real-world pattern

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)