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Inductive Reasoning 1 2 3 4 5 6 … n … 20 0 3 10 21 36 55 … ? … ?
Linear Sequences 1, 3, 5, 7, 9, … 2, 4, 6, 8, 10, … 3, 8, 13, 18, 23, … 7, 7, 7, 7, 7, …
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 …
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 … x y # of term Term  Value
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.
1  2  3  4  5 … 4  5  6  7  8 … x y # 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.
1  2  3  4  5 … 4  5  6  7  8 … x y # of term Term  Value Let’s look at the change in each term or gap between terms. 1 1 1 1 Note that the change in y is 1.  The change in x is also 1.
1  2  3  4  5 … 4  5  6  7  8 … x y # of term Term  Value 1 1 1 1 Since the sequence is linear it has the following form: Y = mX + b m = 1 or the gap between terms.
1  2  3  4  5 … 4  5  6  7  8 … x y # of term Term  Value 1 1 1 1 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.
1  2  3  4  5 … 4  5  6  7  8 … x y # of term Term  Value 1 1 1 1 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
1  2  3  4  5 … 4  5  6  7  6 … x y # of term Term  Value 1 1 1 1 It works.  Look at each term. The rule is  y = x + 3 5 = 2 + 3 6 = 3 + 3 7 = 4 + 3 8 = 5 + 3
1 more time. That was the first try.  Let’s do another. 1  2  3  4  5 … 6  11 16 21 26 … x y # of term Term  Value 5 5 5 5 Since each sequence can be thought of and viewed as an ordered pair, they can be graphed.
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.
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 … x y # of term Term  Value 5 5 5 5
1  2  3  4  5 … 6  11 16 21 26 … x y # of term Term  Value 5 5 5 5 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.
1  2  3  4  5 … 6  11 16 21 26 … x y # of term Term  Value 5 5 5 5 Since the sequence is linear it has the following form: Y = mX + b m = 5 or the gap between terms.
1  2  3  4  5 … 6  11 16 21 26 … x y # of term Term  Value 5 5 5 5 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.
1  2  3  4  5 … 6  11 16 21 26 … x y # of term Term  Value 5 5 5 5 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.
1  2  3  4  5 … 6  11 16 21 26 … x y # of term Term  Value 5 5 5 5 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
Let’s see if we can do these quickly 1  2  3  4  5 …  x 9  13 17 21 25 … x y # of term Term  Value 4 4 4 4 y = mx + b y = 4x + b 9 = 4(1) +b 5 = b y = 4x + 5 4x + 5
Again 1  2  3  4  5 …  x 5  8  11 14 17 … x y # of term Term  Value 3 3 3 3 y = mx + b y = 3x + b 5 = 3(1) +b 2 = b y = 3x + 2 3x + 2
And Again 1  2  3  4  5 …  x 4  11 18 25 32 … x y # of term Term  Value 7 7 7 7 y = mx + b y = 7x + b 4 = 7(1) +b -3 = b y = 7x - 3 7x - 3
One more time ! 1  2  3  4  5 …  x 8  15 22 29 36 … x y # of term Term  Value 7 7 7 7 y = mx + b y = 7x + b 8 = 7(1) +b 1 = b y = 7x + 1 7x + 1
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.
Let’s try a few more ! 1  2  3  4  5 …  x 5  7  9  11 13 … x y # of term Term  Value 2 2 2 2 y = mx + b y = 2x + b 5 = 2(1) +b 3 = b y = 2x + 3 2x + 3
And Another ! 1  2  3  4  5 …  x 8  18  28   38   48 … x y # of term Term  Value 10 10 10 10 y = mx + b y = 10x + b 8 = 10(1) +b -2 = b y = 10x - 2 10x -2
Let’s try tricky ones ! 1  2  3  4  5 …  x 7  7  7   7   7 … x y # of term Term  Value 0 0 0 0 y = mx + b y = 0x + b 7 = 0(1) +b 7 = b y =  7 7
And Another ! 1  2  3  4  5 …  x 17  14   11   8  5  … x y # of term Term  Value -3 -3 -3 -3 y = mx + b y = -3x + b 17 = -3(1) +b 20 = b y = -3x + 20 -3x + 20
Let’s find the 20 th   Term     3 13 18 23 28 8 5 5 5 5 y = 5x + b 3 = 5(1) + b -2 = + b 5x - 2 5(20) - 2 98 x 1 2 3 4 5 6   … x   … 20 y                     GAP            
Again     3 7 9 11 13 5 2 2 2 2 y = 2x + b 3 = 2(1) + b 1 = + b 2x + 1 2(20) + 1 41 x 1 2 3 4 5 6   … x   … 20 y                     GAP            
And Again     3 11 15 19 23 7 4 4 4 4 y = 4x + b 3 = 4(1) + b - 1 = + b 4x - 1 4(20) - 1 79 x 1 2 3 4 5 6   … x   … 20 y                     GAP            
One Last Time     3 13 18 23 28 8 5 5 5 5 y = 5x + b 3 = 5(1) + b - 2 = + b 5x - 2 5(20) - 2 98 x 1 2 3 4 5 6   … x   … 20 y                     GAP            
C’est fini. Good day and good luck. A Senior Citizen Production That’s all folks.

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2LinearSequences

  • 1. Inductive Reasoning 1 2 3 4 5 6 … n … 20 0 3 10 21 36 55 … ? … ?
  • 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 … x y # 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 … x y # 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 … x y # of term Term Value Let’s look at the change in each term or gap between terms. 1 1 1 1 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 … x y # of term Term Value 1 1 1 1 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 … x y # of term Term Value 1 1 1 1 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 … x y # of term Term Value 1 1 1 1 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 … x y # of term Term Value 1 1 1 1 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. 1 more time. That was the first try. Let’s do another. 1 2 3 4 5 … 6 11 16 21 26 … x y # of term Term Value 5 5 5 5 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 … x y # of term Term Value 5 5 5 5
  • 15. 1 2 3 4 5 … 6 11 16 21 26 … x y # of term Term Value 5 5 5 5 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 … x y # of term Term Value 5 5 5 5 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 … x y # of term Term Value 5 5 5 5 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 … x y # of term Term Value 5 5 5 5 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 … x y # of term Term Value 5 5 5 5 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 … x y # of term Term Value 4 4 4 4 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 … x y # of term Term Value 3 3 3 3 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 … x y # of term Term Value 7 7 7 7 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 … x y # of term Term Value 7 7 7 7 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 … x y # of term Term Value 2 2 2 2 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 … x y # of term Term Value 10 10 10 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 … x y # of term Term Value 0 0 0 0 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 … x y # of term Term Value -3 -3 -3 -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     3 13 18 23 28 8 5 5 5 5 y = 5x + b 3 = 5(1) + b -2 = + b 5x - 2 5(20) - 2 98 x 1 2 3 4 5 6   … x   … 20 y                     GAP            
  • 30. Again     3 7 9 11 13 5 2 2 2 2 y = 2x + b 3 = 2(1) + b 1 = + b 2x + 1 2(20) + 1 41 x 1 2 3 4 5 6   … x   … 20 y                     GAP            
  • 31. And Again     3 11 15 19 23 7 4 4 4 4 y = 4x + b 3 = 4(1) + b - 1 = + b 4x - 1 4(20) - 1 79 x 1 2 3 4 5 6   … x   … 20 y                     GAP            
  • 32. One Last Time     3 13 18 23 28 8 5 5 5 5 y = 5x + b 3 = 5(1) + b - 2 = + b 5x - 2 5(20) - 2 98 x 1 2 3 4 5 6   … x   … 20 y                     GAP            
  • 33. C’est fini. Good day and good luck. A Senior Citizen Production That’s all folks.

Notas do Editor

  1. 4, 5, 6, 7, 6
  2. 4, 5, 6, 7, 6