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Arithmetic sequences have a common difference, which is the same as them having a constant rate of change. Thus, even though arithmetic sequences have domains that are discrete, they are similar to linear functions. Just as linear function can be described and analyzed using function rules, so can arithmetic sequences.

An explicit rule of a sequence is a function rule where the input is a term's position, and the output is the value of the term. This is opposed to recursive rules, where a term's value is found using previous terms. Explicit rules are often written as $a_{n}=f(n),$ where $f(n)$ is the function rule. An example would be the sequence $2,3,4,…,$ which can be described by the explicit rule $a_{n}=n+1,$

where $n$ is the position of a term in the sequence. It says that the first term, $a_{1},$ has the value $2,$ the second term, $a_{2},$ has the value $3,$ and so on.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.

$a_{n}=a_{1}+(n−1)d$

$a_{n}=4+(n−1)3$

DistrDistribute $3$

$a_{n}=4+3n−3$

SubTermSubtract 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$n=12$

$a_{12}=1+3⋅12$

MultiplyMultiply

$a_{12}=1+36$

AddTermsAdd terms

$a_{12}=37$

The twelfth term is $37.$

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.

$a_{n}=a_{1}+(n−1)d$

$a_{n}=1+(n−1)2$

DistrDistribute $2$

$a_{n}=1+2n−2$

SubTermSubtract term

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

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

Substitute$a_{n}=53$

$53=-1+2n$

AddEqn$LHS+1=RHS+1$

$54=2n$

DivEqn$LHS/2=RHS/2$

$27=n$

RearrangeEqnRearrange equation

$n=27$

Row $27$ has $53$ pellets.

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 $d.$

$a_{6}−a_{3}=3d$

$0−15=3d$

SubTermSubtract term

$-15=3d$

DivEqn$LHS/3=RHS/3$

$-5=d$

RearrangeEqnRearrange 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.

$a_{n}=a_{1}+(n−1)d$

$a_{n}=25+(n−1)(-5)$

DistrDistribute $-5$

$a_{n}=25−5n+5$

AddTermsAdd 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 $25,20,15,10,5,and0.$

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