# Math Journal 9-5 Evaluate Simplify

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Math Journal 9-5 Evaluate Simplify
1) 15 β (-13) = 2) 14 π 2 β3 π 2 + π 2 β7π Find the next 4 terms of the Arithmetic Sequence. 3) 7, 4, 1, -2, Find the next 3 terms of the Recursive Sequence. 4) 1, 3, 4, 7, 11,

Unit 2 Day 4: Sequences as Functions
Essential Questions: How can any term of an arithmetic sequence be determined? How do we represent a sequence in function notation?

Patterns in Arithmetic Sequences
Patterns can be thought of asΒ sequences, or a list of numbers. The below example is what type of sequence? Arithmetic 1, 2 , 3 , β¦ +1 Example: The set of Natural Numbers

8 - 3 = +5 Example: 3, 8, 13, 18, β¦ What is the pattern?
We can write an arithmetic sequence recursively if we know the pattern (or rule), and the first term. Writing a sequences recursively helps us find any term in the sequence. Example: 3, 8, 13, 18, β¦ What is the pattern? We can use recursion to find the common difference without βguessingβ or βanalyzingβ. Current term: 8 Previous term: 3 Subtract 3 from the current term: 8 - 3 = +5 +5 is called the common difference.

Example 1 Describe the sequence recursively : 12, 18, 24, 30, β¦
First, find the common difference, label it d. d = 18 β 12 = 6 Now that we determined that this is an arithmetic sequence with a common difference between successive terms, we can predict the following terms: Term # Term 1 12 2 18 3 24 4 30 5 6 You may want to point out that we called the initial term A(1) but we could have defined it as A(0) just changes the formula a little bit. 36 42

Arithmetic nth Formula (nth term):
Is This Always Useful? What are some drawbacks? What if we want to find the 100th term in the sequence? We would have to find all 99 terms that precede it! Arithmetic nth Formula (nth term): an = a1 + d(n - 1) Term I want NOW! 1st Term in the Sequence Common Difference Term Number

an = 6 + (-2)(n - 1) Example 2 an = a1 + d(n - 1) d = 4 - 6 = -2
Use the formula for the following arithmetic sequence, then find the 10th term: 6, 4, 2, 0, β¦ d = = -2 a1 = 6 n = 10 an = 6 + (-2)(n - 1) a10 = 6 - 2(10 - 1) a10 = 6 - 2(9) a10 = a10 = -12

π π = π 1 +π πβ1 π 1 =3, π=6, π=9 an = 3 + 6(9 - 1) a9 = 51 Example 3
Use the arithmetic formula to determine the 9th term in the sequence: 3, 9, 15, 21, β¦ π π = π 1 +π πβ1 π 1 =3, π=6, π=9 an = 3 + 6(9 - 1) a9 = 51 Could ask if they could do it another way, graph the points, make a table of input v output

Writing An Arithmetic Sequence as a Function
1. List the given sequence. 2. Write down the formula: π π = π 1 +π πβ1 .3. Identify the first term: π 1 = 4. Calculate the common difference: π= 5. Plug π 1 and π into the Arithmetic nth Formula. 6. Distribute the π value. 7. Combine all like terms if needed. 8. Change the π π to function notation a(n).

Writing An Arithmetic Sequence as a Function
Consider the sequence 7, 11, 15, 19, β¦ Think of each term as the output of a function. Think of the term number (n) as the input. Term number (n) 1 2 3 4 input Term 7 11 15 19 output

Example 4 Following the Steps!! 7, 11, 15, 19 π π = π 1 +π πβ1 π 1 = 7
Term # 1 2 3 4 input Term 7 11 15 19 output Following the Steps!! 7, 11, 15, 19 π π = π 1 +π πβ1 π 1 = 7 π = (11 - 7) = 4 an = 7 + 4(n - 1) an = 7 + 4n β 4 an = 4n + 3 a(n) = 4n + 3

an = 6 + -2(n - 1) an= 6 + -2n + 2 an= -2n + 8 a(n) = -2n + 8
Example 5 Write the arithmetic sequence as a function. 6, 4, 2, 0, β¦ π π = π π +π(πβπ) an = (n - 1) an= n + 2 an= -2n + 8 a(n) = -2n + 8

an = 3 + 6(n - 1) an= 3 + 6n - 6 an= 6n - 3 a(n) = 6n - 3 Example 6
Write the function for the arithmetic sequence. 3, 9, 15, 21, β¦ π π = π π +π(πβπ) an = 3 + 6(n - 1) an= 3 + 6n - 6 an= 6n - 3 a(n) = 6n - 3

Summary Essential Questions: How can any term of an arithmetic sequence be determined? How do we represent a sequence in function notation? Take 1 minute to write 2 sentences answering the essential questions.