1. Inductive Reasoning 123456…n…20 0310213655…?…?

2. Linear Sequences 1, 3, 5, 7, 9, … 2, 4, 6, 8, 10, … 3, 8, 13, 18, 23, … 7, 7, 7, 7, 7, …

3. Sequences Terms: 4, 5, 6, 7, 8 … Each number is a term of the sequence. Each term of the sequence is associated with the counting numbers. The counting number represent the terms location: First, second, third, etc. 1 2 3 4 5 … 4 5 6 7 6 …

4. Since each sequence can be thought of and viewed as an ordered pair, they can be graphed. 1 2 3 4 5 … 4 5 6 7 8 … 1 2 3 4 5 … 4 5 6 7 8 … xyxy # of term Term Value

5. As you can see the sequence is a line of integer values. Hence we call it a linear sequence. We can find the succeeding points by graphing or just visually recognizing the pattern. However, graphing is very time consuming.

6. 1 2 3 4 5 … 4 5 6 7 8 … xyxy # of term Term Value Recognizing the pattern is not efficient for finding the 50 th term because you need to find the first 49 terms to compute the 50 th term. Therefore, it would be quicker if we could come up with a simple algebraic rule.

7. 1 2 3 4 5 … 4 5 6 7 8 … xyxy # of term Term Value Let’s look at the change in each term or gap between terms. 1111 Note that the change in y is 1. The change in x is also 1.

8. 1 2 3 4 5 … 4 5 6 7 8 … xyxy # of term Term Value 1111 Since the sequence is linear it has the following form: Y = mX + b m = 1 or the gap between terms.

9. 1 2 3 4 5 … 4 5 6 7 8 … xyxy # of term Term Value 1111 Since the sequence is linear it has the following form: Y = mX + b m = 1 or the gap between terms. The slope will always be the gap between terms because the change in x will always be 1.

10. 1 2 3 4 5 … 4 5 6 7 8 … xyxy # of term Term Value 1111 Y = (1)X + b To find the value of b, use the first term and substitute 1 for x and substitute 4 for y. 4 = (1)(1) + b 4 = 1 + b 3 = b The rule is y = x + 3

11. 1 2 3 4 5 … 4 5 6 7 6 … xyxy # of term Term Value 1111 It works. Look at each term. The rule is y = x + 3 5 = 2 + 3 6 = 3 + 3 7 = 4 + 3 8 = 5 + 3

12. That was the first try. Let’s do another. 1 2 3 4 5 … 6 11 16 21 26 … xyxy # of term Term Value 5555 1 more time. Since each sequence can be thought of and viewed as an ordered pair, they can be graphed.

13. As you can see the sequence is a line of integer values. Hence we call it a linear sequence. We can find the succeeding points by graphing or just visually recognizing the pattern. However, graphing is very time consuming.

14. Recognizing the pattern is not efficient for finding the 50 th term because you need to find the first 49 terms to compute the 50 th term. Therefore, it would be quicker if we could come up with a simple algebraic rule. 1 2 3 4 5 … 6 11 16 21 26 … xyxy # of term Term Value 5555

15. 1 2 3 4 5 … 6 11 16 21 26 … xyxy # of term Term Value 5555 Let’s look at the change in each term or gap between terms. Note that the change in y is 5. The change in x is also 1.

16. 1 2 3 4 5 … 6 11 16 21 26 … xyxy # of term Term Value 5555 Since the sequence is linear it has the following form: Y = mX + b m = 5 or the gap between terms.

17. 1 2 3 4 5 … 6 11 16 21 26 … xyxy # of term Term Value 5555 Since the sequence is linear it has the following form: Y = mX + b m = 1 or the gap between terms. The slope will always be the gap between terms because the change in x will always be 1.

18. 1 2 3 4 5 … 6 11 16 21 26 … xyxy # of term Term Value 5555 Y = 5X + b To find the value of b, use the first term and substitute 1 for x and substitute 6 for y. 6 = 5(1) + b 6 = 5 + b 1 = b The rule is y = 5x + 1 This is usually done mentally by multiplying the term # by the gap and figuring out what else is needed to make the term value.

19. 1 2 3 4 5 … 6 11 16 21 26 … xyxy # of term Term Value 5555 The rule is y = 5x + 1 11 = 5(2) + 1 It works. Look at each term. 16 = 5(3) + 1 21 = 5(4) + 1 26 = 5(5) + 1

20. Let’s see if we can do these quickly 1 2 3 4 5 … x 9 13 17 21 25 … xyxy # of term Term Value 4444 y = mx + b y = 4x + b 9 = 4(1) +b 5 = b y = 4x + 5 4x + 5

21. Again 1 2 3 4 5 … x 5 8 11 14 17 … xyxy # of term Term Value 3333 y = mx + b y = 3x + b 5 = 3(1) +b 2 = b y = 3x + 2 3x + 2

22. And Again 1 2 3 4 5 … x 4 11 18 25 32 … xyxy # of term Term Value 7777 y = mx + b y = 7x + b 4 = 7(1) +b -3 = b y = 7x - 3 7x - 3

23. One more time ! 1 2 3 4 5 … x 8 15 22 29 36 … xyxy # of term Term Value 7777 y = mx + b y = 7x + b 8 = 7(1) +b 1 = b y = 7x + 1 7x + 1

24. Linear sequences must be done quickly. The speed should be almost as fast as you can write. The rule is in the form of m x + b. where x is the number of the term.

25. Let’s try a few more ! 1 2 3 4 5 … x 5 7 9 11 13 … xyxy # of term Term Value 2222 y = mx + b y = 2x + b 5 = 2(1) +b 3 = b y = 2x + 3 2x + 3

26. And Another ! 1 2 3 4 5 … x 8 18 28 38 48 … xyxy # of term Term Value 10 y = mx + b y = 10x + b 8 = 10(1) +b -2 = b y = 10x - 2 10x -2

27. Let’s try tricky ones ! 1 2 3 4 5 … x 7 7 7 7 7 … xyxy # of term Term Value 0000 y = mx + b y = 0x + b 7 = 0(1) +b 7 = b y = 7 7

28. And Another ! 1 2 3 4 5 … x 17 14 11 8 5 … xyxy # of term Term Value -3 y = mx + b y = -3x + b 17 = -3(1) +b 20 = b y = -3x + 20 -3x + 20

29. Let’s find the 20 th Term x123456 … x … 20 y GAP 31318 2328 8 5 555 y = 5x + b 3 = 5(1) + b -2 = + b 5x - 2 5(20) - 2 98

30. x123456 … x … 20 y GAP 3 7 9 1113 5 2 222 Again y = 2x + b 3 = 2(1) + b 1 = + b 2x + 1 2(20) + 1 41

31. x123456 … x … 20 y GAP 3 11 15 1923 7 4 444 And Again y = 4x + b 3 = 4(1) + b - 1 = + b 4x - 1 4(20) - 1 79

32. One Last Time x123456 … x … 20 y GAP 3 13 18 2328 8 5 555 y = 5x + b 3 = 5(1) + b - 2 = + b 5x - 2 5(20) - 2 98

33. C’est fini. Good day and good luck. A Senior Citizen Production That’s all folks.