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Writing Recursive Rules for Arithmetic Sequences

There are different ways to express the rules of sequences depending on the information that's known. Explicit rules require the first term, a1,a_1, the common difference, dd and the position of the desired term n.n. Recursive rules have different requirements.

Recursive Rule

A recursive rule gives the first term or terms of a sequence and describes how each term is related to the preceding term(s) with a recursive equation. For example, arithmetic and geometric sequences can be described recursively.

Example arithmetic sequence Example geometric sequence
geometric sequence
geometric sequence
Recursive equation
a1=3,   an ⁣= ⁣an1 ⁣+ ⁣2\begin{gathered} a_1=3 , \ \ \ a_n \! = \! a_{n-1}\! + \! 2 \end{gathered} a1=5,   an ⁣= ⁣an1 ⁣× ⁣2\begin{gathered} a_1=5 , \ \ \ a_n \! = \! a_{n-1}\! \times \! 2 \end{gathered}
One particularly well-known sequence that is defined recursively is the Fibonacci sequence, in which each term is the sum of the two previous terms.

Writing a Recursive Rule for an Arithmetic Sequence

The recursive rule for an arithmetic sequence can be expressed as an=an1+d a_n=a_{n-1}+d where an1a_{n-1} and ana_n are consecutive terms and dd is the common difference. To write the rule, dd must be found. Consider the arithmetic sequence 5,11,17,23, 5,\, 11,\, 17,\, 23,\, \ldots To find d,d, we can subtract any term from the term that comes after it. Let's use a2=11a_2={\color{#0000FF}{11}} and a1=5.a_1={\color{#009600}{5}}. d=a2a1=115=6 d=a_2-a_1={\color{#0000FF}{11}}-{\color{#009600}{5}}=6 Using d=6,d=6, the recursive rule can be written as an=an1+6. a_n=a_{n-1}+6. As it's written, this rule describes any sequence with a common difference of 6.6. For example, it can describe -45,-39,-33,-27,\text{-} 45,\,\text{-} 39,\,\text{-} 33,\,\text{-}27, \ldots as well as 111,117,123,129,135,111, 117, 123, 129, 135, \ldots To ensure the recursive rule defines the given sequence, it is necessary to also give the first term, a1.a_1. Thus, the recursive rule for the given sequence is a1=5an=an1+6.\begin{aligned} &a_1=5 \\ &a_n=a_{n-1}+6. \end{aligned} Now that the recursive rule is known, it can be used to find any term, provided that the previous term is given. For example, since a4=23,a_4=23, a5a_5 can be found. an=an1+6a5=a4+6a5=23+6a5=29\begin{aligned} a_n&=a_{n-1}+6 \\ a_5&=a_4+6 \\ a_5&= 23+ 6 \\ a_5&=29 \end{aligned}

Thus, the 5th5\text{th} term in the sequence is 29.29.

Bonnie recently purchased a used car. Each weekend Bonnie records the odometer of her car. The first four weekends she noted the following miles. 26247263112637526439\begin{aligned} 26\,247 && 26\,311 \\ 26\,375 && 26\,439 \end{aligned}

Create a recursive rule that describes the sequence. Then use the recursive rule to find the odometer's reading the following two weeks, assuming that Bonnie drives exactly the same distance every week.

Show Solution

Before we write the recursive rule for the sequence, let's ensure it's arithmetic.

Notice that Bonnie drives 6464 miles each week. Therefore, the odometer readings form an arithmetic sequence with a common difference of d=64.d=64. Using dd and the given information we can write the recursive rule as follows. a1=26247an=an1+64\begin{aligned} a_1&=26\,247 \\ a_n&=a_{n-1}+64 \end{aligned} We can use the recursive rule to find the odometer readings for the next two weeks, a5a_5 and a6.a_6. We'll use a4=26439a_4=26\,439 to find a5.a_5. an=an1+64a5=a4+64a5=26439+64a5=26503\begin{aligned} a_n&=a_{n-1}+64 \\ a_5&=a_4+64 \\ a_5&=26\,439+64 \\ a_5&=26\,503 \end{aligned} Similarly, we can add 6464 to a5a_5 to find the odometer reading of the next weekend. a6=a5+64=26503+64=26567 a_6=a_5+64=26\,503+64=26\,567 Thus, after the fifth weekend, the odometer will read 2650326\,503 miles and after the sixth weekend, it will read 2656726\,567 miles.


Translating between Explicit and Recursive Rules

As has been explored, it is possible to express sequences with explicit rules and recursive rules. Since the rules present different information about the sequence, it can be useful to translate between the two.

An arithmetic sequence is given by the following explicit rule. Translate it into a recursive rule. an=4+(n1)6. a_n=4+(n-1)6.

Show Solution

A recursive rule for an arithmetic sequence is written as a1=an=an1+d,\begin{aligned} a_1&= \ldots \\ a_n&=a_{n-1}+d, \end{aligned}

where a1a_1 is the first term and dd is the common difference. Since an explicit rule for an arithmetic sequence is written as an=a1+(n1)d,a_n=a_1+(n-1)d, it can be seen that a1=4a_1=4 and d=6.d=6. Thus, we can write the recursive rule for this sequence as a1=4an=an1+6.\begin{aligned} a_1&=4\\ a_n&=a_{n-1}+6. \end{aligned}


An arithmetic sequence is defined by the following recursive rule. Write the explicit rule for the same sequence. a1=-5an=an1+7 \begin{array}{l}a_1=\text{-} 5 \\ a_n=a_{n-1}+7 \end{array}

Show Solution

The explicit rule of an arithmetic sequence is written as an=a1+(n1)d, a_n=a_1+(n-1)d, where a1a_1 is the first term and dd is the common difference. To write an explicit rule, both a1a_1 and dd are needed. From the recursive rule a1=-5an=an1+7\begin{aligned} a_1&=\text{-} 5\\ a_n&=a_{n-1}+7 \end{aligned} it can be seen that a1=-5a_1=\text{-} 5 and d=7.d=7. Thus, the explicit rule can be written and simplified as follows. an=-5+(n1)7an=-12+7n a_n=\text{-} 5+(n-1)7 \quad \Rightarrow \quad a_n=\text{-}12 +7n

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