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{{ printedBook.courseTrack.name }} {{ printedBook.name }} A sequence is a finite or infinite ordered list of numbers, also called terms. An example of a sequence is $1,4,7,10,13,…$

The dots indicate that the sequence continues infinitely, following the same pattern. In this sequence, the difference between each term is $3.$ Any sequence where the difference between consecutive terms is constant, is anFor an arithmetic sequence, the difference between consecutive terms is constant. Meaning, the difference between the first and second term is the same as the difference between the second and the third term, and so forth. This difference is called the common difference and is usually denoted with $d.$ An example of an arithmetic sequence is the following.

Here, the common difference is $d=2.$Is the following sequence arithmetic? If so, write the next three terms in the sequence. $45,39,33,27,21,…$

Show Solution

To determine if the sequence is arithmetic, we have to show that the difference between consecutive terms is constant. The difference between the first and second term is
$39−45=-6.$
The difference is negative, meaning that the second term is $6$ **less** than the first. If this is an arithmetic sequence the next term should be
$39−6=33,$
and it is! Furthermore, $27$ is $6$ less than $33,$ and $21$ is $6$ less than $27.$ Thus, the sequence is arithmetic. To find the next term we subtract $6$ from $21:$
$21−6=15.$
The next term is $15.$ The one after that is $15−6=9,$ and the one after that is $9−6=3.$ To summarize, the sequence is arithmetic, and the next three terms are $15,$ $9,$ and $3.$

In a theater, there are $10$ rows of seats. The first row has $11$ seats, and for each subsequent row, the number of seats increases by $3.$ Write an arithmetic sequence to represent the number of seats in each row, then graph the sequence.

Show Solution

To begin, we can start with the first row. It is given that this row has $11$ seats. Since each row has $3$ more seats than the previous row, we know the second row has $14$ seats. $11+3=14.$ Similarly, the third row has $17$ seats, because $14+3=17.$ Continuing this pattern for the remaining rows, we can write the following sequence. $11,14,17,20,23,26,29,32,35,38.$ To graph the sequence, we can let $x$ be the row number and $y$ be the number of seats in the row. Arranging the sequence in a table can help us see the points that need to be graphed.

Lastly, to graph the sequence, we can plot the points from the table. Notice that the points aren't connected. This is because the row number and the number of seats in each row both have to be whole numbers.

The domain of the sequence is $D={1,2,3,4,5,6,7,8,9,10},$ and the range is $R={11,14,17,20,23,26,29,32,35,38}.$

Consider the arithmetic sequence $1,2.5,4,5.5,7,…$ Here, the common difference is $1.5,$ and the terms can be illustrated in a table, where $n$ represents the term number and $a_{n}$ represents the term.

Because their terms change by a constant amount, arithmetic sequences show a linear relationship. Here, the common difference $d=1.5$ can be considered the slope of the line. In fact, when plotting an arithmetic sequence in a coordinate plane, it resembles the graph of a linear function.

By making this comparison, arithmetic sequences can be considered functions. They have the same characteristics as linear functions, but where linear functions are continuous, both the domain and range for arithmetic sequences are discrete.All arithmetic sequences have some common difference, $d.$ Using this common difference, and the value of the first term, $a_{1},$ it's possible to find an explicit rule that describes the sequence. By thinking of the terms in a sequence using $a_{1}$ and $d,$ a pattern emerges.

$n$ | $a_{n}$ | Using $a_{1}$ and $d$ |
---|---|---|

$1$ | $a_{1}$ | $a_{1}+0d$ |

$2$ | $a_{2}$ | $a_{1}+1d$ |

$3$ | $a_{3}$ | $a_{1}+2d$ |

$4$ | $a_{4}$ | $a_{1}+3d$ |

When $n$ increases by $1,$ the coefficient of $d$ increases by $1$ as well. Due to this, and that the coefficient is $0$ when $n$ is $1,$ the coefficient is always $1$ less than $n.$ Expressing this in a general form gives the explicit rule.

$a_{n}=a_{1}+(n−1)d$

Thus, knowing $a_{1}$ and $d$ is enough to write the explicit rule of an arithmetic sequence. Note that an arithmetic sequence is a linear function where the domain is the positive integers. The difference $d$ is then the slope, and $(1,a_{1})$ is a point on the graph. Substituting this into the point-slope form is an alternative way of finding the rule: $a_{n}−a_{1}=d(n−1).$

This equality can be rearranged into the explicit rule previously stated.The first five terms of an arithmetic sequence are $4,7,10,13,and16.$ Find an explicit rule describing the arithmetic sequence. Then, use the rule to find the twelfth term of the sequence.

Show Solution

To find the rule, we first need to find the common difference, $d,$ of the sequence. We can do this by subtracting one term from the next: $d=7−4=3.$ We also know that the first term of the sequence is $4.$ Substituting these pieces of information into the general form of the explicit rule gives the desired rule.

$a_{n}=a_{1}+(n−1)d$

SubstituteII

$a_{1}=4$, $d=3$

$a_{n}=4+(n−1)3$

Distr

Distribute $3$

$a_{n}=4+3n−3$

SubTerm

Subtract term

$a_{n}=1+3n$

To find the twelfth term of the sequence, we can now substitute $n=12$ into the rule.

$a_{n}=1+3n$

Substitute

$n=12$

$a_{12}=1+3⋅12$

Multiply

Multiply

$a_{12}=1+36$

AddTerms

Add terms

$a_{12}=37$

The twelfth term is $37.$

For an arithmetic sequence, $a_{3}=15anda_{6}=0.$ Write an explicit rule of the sequence and give its first six terms.

Show Solution

To begin, we must determine the common difference, $d.$ Since we do not know the values of two consecutive terms, we cannot directly find $d.$ However, the terms $a_{3}$ and $a_{6}$ are $3$ positions apart, so they must differ by $3d.$

This gives the equation $a_{6}−a_{3}=3d,$ which we can solve for $d.$

$a_{6}−a_{3}=3d$

SubstituteII

$a_{6}=0$, $a_{3}=15$

$0−15=3d$

SubTerm

Subtract term

$-15=3d$

DivEqn

$LHS/3=RHS/3$

$-5=d$

RearrangeEqn

Rearrange equation

$d=-5$

Now that we know the common difference, we have to find the first term of the sequence, $a_{1},$ to write the explicit rule. Using $a_{3}$ and $d,$ we can find $a_{1}.$ Knowing one term, a subsequent term can be found by adding $d.$ Similarly, a previous term is found by subtracting $d.$ $a_{2}=a_{3}−d⇒a_{2}=15−(-5)=20$ Repeating this, we find $a_{1}.$ $a_{1}=a_{2}−d⇒a_{1}=20−(-5)=25$ We now know both $a_{1}$ and $d,$ so we can find the explicit rule.

$a_{n}=a_{1}+(n−1)d$

SubstituteII

$a_{1}=25$, $d=-5$

$a_{n}=25+(n−1)(-5)$

Distr

Distribute $-5$

$a_{n}=25−5n+5$

AddTerms

Add terms

$a_{n}=30−5n$

Thus, the explicit rule is $a_{n}=30−5n.$ We have already found the terms $a_{1},$ $a_{2},$ $a_{3},$ and $a_{6}.$ Finding the last two can be done either by using the explicit rule, or by adding $d$ to $a_{3}$ and then to $a_{4}.$ For simplicity's sake, let's add $d$ to $a_{3}$ to find $a_{4}.$ $a_{4}=a_{3}+d⇒a_{4}=15+(-5)=10$ Then, $a_{5}$ is found using $a_{4}.$ $a_{5}=a_{4}+d⇒a_{5}=10+(-5)=5$ Thus, the first six terms of the sequence are $25,20,15,10,5,and0.$

Pelle is an avid collector of pellets. During his spare time, he likes to arrange his pellets in different patterns. Today, he's chosen to place them in the shape of a triangle. The top row consists of one pellet, the second of three pellets, the third of five pellets, and so on. Write a rule $a_{n}=f(n),$ where $a_{n}$ is the amount of pellets in row $n.$ Then, use the rule to find which row has $53$ pellets.

Show Solution

To begin, we can make sense of the given information. For every row, the amount of pellets increase by $2.$ Thus, we know that $d=2.$ It is also given that the first row consists of $1$ pellet, $a_{1}=1.$ Using this information, we can find the rule.

$a_{n}=a_{1}+(n−1)d$

SubstituteII

$a_{1}=1$, $d=2$

$a_{n}=1+(n−1)2$

Distr

Distribute $2$

$a_{n}=1+2n−2$

SubTerm

Subtract term

$a_{n}=-1+2n$

$a_{n}=-1+2n$

Substitute

$a_{n}=53$

$53=-1+2n$

AddEqn

$LHS+1=RHS+1$

$54=2n$

DivEqn

$LHS/2=RHS/2$

$27=n$

RearrangeEqn

Rearrange equation

$n=27$

Row $27$ has $53$ pellets.

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