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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
where f(n) is the function rule. An example would be the sequence $2,3,4,…,$ which can be described by the explicit rule
where n is the position of a term in the sequence. It says that the first term, a1, has the value 2, the second term, a2, 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, a1, it's possible to find an explicit rule that describes the sequence. By thinking of the terms in a sequence using a1 and d, a pattern emerges.

n | $a_{n}$ | Using a1 and d |
---|---|---|

1 | a1 | a1+0d |

2 | a2 | a1+1d |

3 | a3 | a1+2d |

4 | a4 | a1+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.

The first five terms of an arithmetic sequence are

4,7,10,13,and 16.

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

a1=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.

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, a1=1. Using this information, we can find the rule.

$a_{n}=a_{1}+(n−1)d$

SubstituteII

a1=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.

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 a3 and a6 are 3 positions apart, so they must differ by 3d.

This gives the equationa6−a3=3d,

which we can solve for d.
a6−a3=3d

SubstituteII

a6=0, a3=15

0−15=3d

SubTerm

Subtract term

-15=3d

DivEqn

$LHS/3=RHS/3$

-5=d

RearrangeEqn

Rearrange equation

d=-5

$a_{n}=a_{1}+(n−1)d$

SubstituteII

a1=25, $d=-5$

$a_{n}=25+(n−1)(-5)$

Distr

Distribute -5

$a_{n}=25−5n+5$

AddTerms

Add terms

$a_{n}=30−5n$

25,20,15,10,5,and 0.

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