Day 28 Arithmetic Sequence. Find the next two terms of each sequence. Then describe the pattern. The equations will be completed later.

Transcription

1 Find the next two terms of each sequence. Then describe the pattern. The equations will be completed later. 1, 3, 5, 7, 9,, Description: Equation: 2, 7, 12, 17, 22,, Description: Equation: -416, -323, -230, -137,, Description: Equation: -2, -5, -8, -11,, Description: Equation: All of the patterns above are called arithmetic sequences. Hopefully, you noticed something about their pattern that makes them similar. Complete the sentence below by writing a description of the pattern you noticed above. Arithmetic sequences are sequences of numbers where. Let s look more closely at the first pattern 1, 3, 5, 7, 9 Suppose the domain is the position of a term (1, 2, 3, 4, etc.) and the range is the term (1, 3, 5, 7, 9, etc.). Make a graph of the points that are made (position, term) with the pattern. What quadrant(s) are these points in? Why? What kind of graph do you have? Write an equation for the graph. How does this equation relate to the graph? How does this equation relate to the pattern? Do you think the graphs of other arithmetic sequences would look similar? Why or why not? Checkpoint 1: Stop at this point for class comparison. If you are done before others, make equations for the other three patterns listed at the top. HighSchoolMathTeachers 2018 Page 1

2 Now, everyone should have the same equation for the pattern 1, 3, 5, 7, 9 However, we have a problem. This equation makes us use a number that is not in our pattern (-1). Let s say we want to use 1 as a starting point instead of -1 (since 1 is our first term in our sequence). So, suppose our equation is now y = 2x + 1. Our new equation y = 2x + 1 makes our pattern shift over one term (one x value). This means we are adding one too many times! Let s alter the equation slightly to y = 2(x 1) + 1. This will shift all the x values (just like we ve done before) and we won t be adding the extra value of d. Now, we have an equation y = 2 (x 1) + 1 that uses the first term and the common difference (slope). This can be used to make an equation for an arithmetic sequence. Let s use d = common difference, a 1 = first term, and a n = nth term. So, the nth term of an arithmetic sequence can be found by a n = a 1 + (n 1) d HighSchoolMathTeachers 2018 Page 2

3 Checkpoint 2: Find the rule/equation for the 2 nd pattern using the formula above. Now that you know arithmetic sequences need a common difference (number added or subtracted to the pattern) and you know how to find the nth term (or equation) for an arithmetic sequence, let s try some problems. Example 1: Is the sequence arithmetic? If so, what s the common difference? If not, why not? A) 2, -3, -8, -13, B) 1, 5/4, 3/2, 7/4, C) a n = n 2 D) a n = 4n + 3 a 1 2 Example 2: Write the first 5 terms if and d = 7. Checkpoint 3: Let s make sure we are on the right track with examples 1 and 2. Example 3: Write the rule/equation for the given information. A) a 1 = 2, d = 3 B) a 1 2, a 2 9 Example 4: Find the indicated term of each arithmetic sequence. First, find the equation, then plug in your n. A) a 1 = -4, d = 6, n = 9 B) a 20 for a 1 = 15, d = -8 Checkpoint 4: Let s make sure we got the answers to examples 3 and 4. HighSchoolMathTeachers 2018 Page 3

4 Example 5: Write the equation for the nth term of each arithmetic sequence. A) 31, 17, 3,. Now, the next two are slightly different. I will give you a term and the d but the term isn t the first one. You need to work backward to find the first term. B) a 7 = 21, d = 5 We know that a n = a 1 + (n 1) d So, using the given information, we have 21 = a 1 + (7 1) 5 Simplify and solve for a 1. Now, find the equation. C) Follow the steps with this information: a 6 =12, d = 8. Checkpoint 5: Did we follow that? Example 6: Find the arithmetic means (missing terms) in each sequence. A) 6,,,, 42 B) 24,,,,, -1 Challenge: Let s do this one together. Use the given information to write an equation that represents the nth term in each arithmetic sequence. The 19 th term of the sequence is 131. The 61 st term is 509. HighSchoolMathTeachers 2018 Page 4

5 Answer Key Find the next two terms of each sequence. Then describe the pattern. The equations will be completed later. 1, 3, 5, 7, 9, 11, 13 Description: Add 2 to the previous term 2, 7, 12, 17, 22, 27, 32 Description: Add 5 to the previous term -416, 323, 230, 137, 44, 49, Description: Add 93 to the previous term -2, 5, 8, 11, 14, 17, All of the patterns above are called arithmetic sequences. Hopefully, you noticed something about their pattern that makes them similar. Complete the sentence below by writing a description of the pattern you noticed above. Arithmetic sequences are sequences of numbers where the difference between one term and the next is a constant. Let s look more closely at the first pattern 1, 3, 5, 7, 9 Suppose the domain is the position of a term (1, 2, 3, 4, etc.) and the range is the term (1, 3, 5, 7, 9, etc.). Make a graph of the points that are made (position, term) with the pattern. What quadrant(s) are these points in? Why? We see all points are in the first quadrant. That s because all domain and range values are positive numbers. What kind of graph do you have? We have a linear graph here. Write an equation for the graph y = 2x 1 How does this equation relate to the graph? How does this equation relate to the pattern? This equation represents the pattern of the sequence. Do you think the graphs of other arithmetic sequences would look similar? Why or why not? Yes, the graphs of other arithmetic sequences would look similar, because all graphs of arithmetic sequences are linear graphs. HighSchoolMathTeachers 2018 Page 5

6 Now, everyone should have the same equation y = 2x 1 for the pattern 1, 3, 5, 7, 9 However, we have a problem. This equation makes us use a number that is not on our pattern (1). Let s say we want to use 1 as a starting point instead of 1 (since 1 is our first term in our sequence). So, suppose our equation is now y = 2x + 1. Our new equation y = 2x + 1 makes our pattern shift over one term (one x value). This means we are adding one too many times! Let s alter the equation slightly to y = 2(x 1) + 1. This will shift all the x values (just like we ve done before) and we won t be adding the extra value of d. We notice it works now Now, we have an equation y = 2(x 1) + 1 that uses the first term and the common difference (slope). This can be used to make an equation for an arithmetic sequence. Let s use d = common difference, a1= first term, and an= nth term. So, the nth term of an arithmetic sequence can be found by a n = a 1 + (n 1)d Checkpoint 2: Find the rule/equation for the 2nd pattern using the formula above. a n = 2 + 5(n 1) Now that you know arithmetic sequences need a common difference (number added or subtracted to the pattern) and you know how to find the nth term (or equation) for an arithmetic sequence, let s try some problems. Example 1: Is the sequence arithmetic? If so, what s the common difference? If not, why not? A) 2, 3, 8, 13 This sequence is arithmetic. Its common difference is 5. B) 1, 5/4, 3/2, 7/4 The sequence is arithmetic. The common difference is ¼. C) a n = n 2 The sequence is not arithmetic, because the common the difference between one term and the next is not a constant. D) a n = 4n + 3 The sequence is arithmetic. The common difference is 4. Example 2: Write the first 5 terms if a and 1 = 2 d = 7 a 1 = 2 a 2 = 5 a 3 = 12 a 4 = 19 a 5 = 26 HighSchoolMathTeachers 2018 Page 6

7 Checkpoint 3: Let s make sure we are on the right track with examples 1 and 2. Example 3: Write the rule/equation for the given information. A) a 1 = 2, d = 3 a n = 2 + 3(n 1) B) a 1 = 2, a 2 = 9 a n = 2 + 7(n 1) Example 4: Find the indicated term of each arithmetic sequence. First find the equation, then plug in your n. A) a 1 = 4, d = 6, n = 9 a n = 4 + 6(n 1) a 9 = 4 + 6(9 1) = 44 B) a 20 for a 1 = 15, d = 8 a n = 15 8(n 1) a 20 = 15 8(20 1) = 137 Checkpoint 4: Let s make sure we got the answers to examples 3 and 4. Example 5: Write the equation for the nth term of each arithmetic sequence. A) 31, 17, 3 a n = 31 14(n 1) Now, the next two are slightly different. I will give you a term and the d but the term isn t the first one. You need to work backward to find the first term. B) a 7 = 21, d = 5 We know that a n = a 1 + (n 1)d So, using the given information, we have 21 = a 1 + (7 1)5 Simplify and solve for a 1. a 1 = 9 Now, find the equation. a n = 9 + 5(n 1) C) Follow the steps with this information: a 6 = 12, d = 8 12 = a 1 + (6 1)8 a 1 = 28 a n = (n 1) Checkpoint 5: Did we follow that? Yes, we did! Example 6: Find the arithmetic means (missing terms) in each sequence. A) 6, 15, 24, 33, 42 B) 24, 19, 14, 9, 4, 1 Challenge: Let s do this one together. Use the given information to write an equation that represents the nth term in each arithmetic sequence. The 19 th term of the sequence is 131. The term is the 61 st 509. a n = 31 + (n 1)9 HighSchoolMathTeachers 2018 Page 7