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a n = a 1 + (n 1) d a 1 First Term in Sequence n the position of - PowerPoint PPT Presentation

Jan 12, 2024 •343 likes •667 views

D AY 28 - A RITHMETIC S EQUENCES A RITHMETIC S EQUENCE An ARITHMETIC SEQUENCE is where the rule of the pattern is always ADDED. The rule is called the COMMON DIFFERENCE A RITHMETIC S EQUENCE You can use the following formula to

  1. D AY 28 - A RITHMETIC S EQUENCES

  2. A RITHMETIC S EQUENCE An ARITHMETIC SEQUENCE is where the ‘rule’ of the pattern is always ADDED. The ‘rule’ is called the COMMON DIFFERENCE

  3. A RITHMETIC S EQUENCE You can use the following formula to find any term in an Arithmetic Sequence: a n = a 1 + (n – 1) d a 1 – First Term in Sequence n – the position of requested term d – the common difference

  4. A RITHMETIC S EQUENCE What is the 19 th term in the following sequence: 101, 93, 85, 77, 69….. Since the common difference is -8, substitute the known values into the formula: a n = a 1 + (n – 1) d a 19 = 101 + (19 - 1) ·(-8)

  5. A RITHMETIC S EQUENCE a 19 = 101 + (19 - 1) ·(-8) = 101+ 18 ·(-8) = 101 + -144 = -43 So, now we know that the 19 th term in the sequence is -43.

  6. Now lets practice: What is the 87 th term (a87) in this sequence? 4, 27, 50, 73, 96…..

  7. F IND THE NEXT TWO TERMS OF EACH SEQUENCE . T HEN DESCRIBE THE PATTERN . T HE EQUATIONS WILL BE COMPLETED LATER . 11 13 1, 3, 5, 7, 9, _____, _____ Description: ____________ Add 2 to the previous term Equation: ___________ 27 32 2, 7, 12, 17, 22, ______, ______ Add 5 to the previous term Description: ____________ Equation: ___________

  8. 44 49 -416, -323, -230, -137, _____, _____ Add 39 to the previous term Description: ____________ Equation: ___________ 14 17 -2, -5, -8, -11, _____, ______ Add 3 to the previous term Description: ____________ Equation: ___________

  9. 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.

  10. 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.

  12. 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 𝑧 = 2𝑦 – 1

  13. 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.

  14. 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.

  15. Now, everyone should have the same equation 𝑧 = 2𝑦 − 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).

  16. So, suppose our equation is now y = 2x + 1. See if this works for the pattern. Try x = 1 (for the first term). Try x = 2 (for the second term). Try x = 3 (for the third term). What do you notice? 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!

  17. Let’s alter the equation slightly to 𝑧 = 2(𝑦 – 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. See if this works for the pattern. Try x = 1 (for the first term). Try x = 2 (for the second term). Try x = 3 (for the third term). What do you notice? We notice it works now

  18. Now, we have an equation 𝑧 = 2(𝑦 − 1) + 1 that uses the first term and the common difference (slope). This can be used to make any equation for any arithmetic sequence. Let’s use d = common difference, a 1 = first term, and a n = nth term. So, the nth term of any arithmetic sequence can be found by 𝑏 𝑜 = 𝑏 1 + (𝑜 − 1)𝑒 Checkpoint 2: Find the rule/equation for the 2nd pattern using the formula above. 𝑏 𝑜 = 2 + 5(𝑜 − 1)

  19. 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 any arithmetic sequence, let’s try some problems.

  20. 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.

  21. C) 𝑏 𝑜 = 𝑜 2 The sequence is not arithmetic, because the common the difference between one term and the next is not a constant. D) 𝑏 𝑜 = 4𝑜 + 3 The sequence is arithmetic. The common difference is 4.

  22. Example 2: Write the first 5 terms if a and 1 = 2 𝑒 = − 7 𝑏 1 = 2 𝑏 2 = −5 𝑏 3 = −12 𝑏 4 = −19 𝑏 5 = −26

  23. Checkpoint 3: Let’s make sure we are on the right track with examples 1 and 2.

  24. E XAMPLE 3: Write the rule/equation for the given information. A) 𝑏 1 = 2, 𝑒 = 3 𝒃 𝒐 = 𝟑 + 𝟒(𝒐 − 𝟐) B) 𝑏 1 = 2, 𝑏 2 = 9 𝒃 𝒐 = 𝟑 + 𝟖(𝒐 − 𝟐)

  25. E XAMPLE 4: Find the indicated term of each arithmetic sequence. First find the equation, then plug in your n. A) 𝑏 1 = −4, 𝑒 = 6, 𝑜 = 9 𝒃 𝒐 = − 𝟓 + 𝟕(𝒐 − 𝟐) 𝒃 𝟘 = − 𝟓 + 𝟕 𝟘 − 𝟐 = 𝟓𝟓 B) 𝑏 20 𝑔𝑝𝑠 𝑏 1 = 15, 𝑒 = −8 𝒃 𝒐 = 𝟐𝟔 − 𝟗 𝒐 − 𝟐 𝒃 𝟑𝟏 = 𝟐𝟔 − 𝟗 𝟑𝟏 − 𝟐 = −𝟐𝟒𝟖

  26. Checkpoint 4: Let’s make sure we got the answers to examples 3 and 4.

  27. E XAMPLE 5: Write the equation for the nth term of each arithmetic sequence. A) 31 , 17, 3… 𝒃 𝒐 = 𝟒𝟐 − 𝟐𝟓(𝒐 − 𝟐) 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 backwards to find the first term.

  28. B) 𝑏 7 = 21, 𝑒 = 5 We know that 𝑏 𝑜 = 𝑏 1 + 𝑜 − 1 𝑒 So, using the given information, we have 21 = 𝑏 1 + 7 − 1 5 Simplify and solve for 𝑏 1 . 𝒃 𝟐 = −𝟘 Now, find the equation. 𝒃 𝒐 = −𝟘 + 𝟔(𝒐 − 𝟐)

  29. C) Follow the steps with this information: 𝑏 6 = 12, 𝑒 = 8 𝟐𝟑 = 𝒃 𝟐 + 𝟕 − 𝟐 𝟗 𝒃 𝟐 = −𝟑𝟗 𝒃 𝒐 = −𝟑𝟗 + 𝟗(𝒐 − 𝟐)

  30. Checkpoint 5: Did we follow that? Yes, we did!

  31. E XAMPLE 6: Find the arithmetic means (missing terms) in each sequence. A) 6, 15, 24, 33, 42 B) 24, 19, 14, 9, 4, 1

  32. C HALLENGE : L ET ’ 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 ‘ th ’ 61 st 509. 𝒃 𝒐 = −𝟒𝟐 + 𝒐 − 𝟐 𝟘

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