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Alg 1B 8-6 Geometric Sequences

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Alg 1B 8-6 Geometric Sequences Powered By Docstoc
					Lesson 8-6
Geometric Sequences,
page 424

   Objectives: To form geometric
   sequences and to use formulas when
   describing geometric sequences.
Why should we learn this?

 One    real-world connection is
   to find the height of a ball
   after a number of bounces.
   (example 5)
REVIEW from Chapter 5

  Sequence – a number pattern
  Term – each number in a sequence
  Arithmetic sequence – add a fixed
   number to each term to find next term
  Common difference – fixed number
   being added to each term
Definitions

    Geometric sequence – multiply a term
     by a fixed number to find next term

    Common ratio – fixed number being
     multiplied in a geometric sequence
Find the common ratio of each sequence.

a. 3, –15, 75, –375, . . .
    3                –15              75           –375

             (–5)           (–5)            (–5)
     The common ratio is –5.
         3       3   3
b. 3, 2 , 4 , 8 , ...
                         3            3                3
     3                   2            4                8

                 1                1                1
                2
                                 2
                                                  2
                                          1
     The common ratio is 2 .


                                                       8-6
Example 1, page 424

 Find the common ratio.    On your own…
 a) 750, 150, 30, 6, …    C) 4, 6, 9, 13.5, …



    THINK, PAIR,
     SHARE…
 b) -3, -6, -12, -24, …
HINT

    When trying to find the common ratio,
     divide a term by the previous term.
     Repeat to find the common ratio.

    Example: -2, 8, -32, 128, …

 8 / -2 = -4           -32 / 8 = -4
 64 / -32 = -4         Common ratio = -4
    Find the next three terms of the sequence 5, –10, 20, –40, . . .

5       –10           20      –40

    (–2)     (–2)        (–2)

The common ratio is –2.

The next three terms are –40(–2) = 80, 80(–2) = –160, and –160(–2) = 320.




                                   8-6
Example 2, page 425
  Find the next two         C) 1.1, 2.2, 4.4, 8.8
   terms in the
   sequence.
  A) 1, 3, 9, 27,…




    B) 120, -60, 30, -
     15,…
Geometric or Arithmetic?

    Is each term being multiplied by fixed number?
     If so, what kind of sequence is it?

    Does each term have a fixed number being
     added? If so, what kind of sequence is it?

    If not, it’s neither geometric or arithmetic.
    Determine whether each sequence is arithmetic
    or geometric.

a. 162, 54, 18, 6, . . .


    62        54           18         6

         1            1          1
      3            3          3


    The sequence has a common ratio.

    The sequence is geometric.




                                     8-6
              (continued)



b. 98, 101, 104, 107, . . .


    98        101        104         107

         +3         +3         +3


    The sequence has a common difference.


    The sequence is arithmetic.



                                    8-6
Example 3, page 425
Arithmetic, Geometric or Neither?

     a) 2, 4, 6, 8,…        c) 1, 3, 5, 7,…




     b) 2, 4, 8, 16, …
USING A FORMULA

            A(n) = a • r n – 1

  A(n) = nth term
  a = first term
  r = common ratio
  n = term number
    Find the first, fifth, and tenth terms of the sequence that has the rule
    A(n) = –3(2)n – 1.


first term: A(1) = –3(2)1 – 1 = –3(2)0 = –3(1) = –3


fifth term: A(5) = –3(2)5 – 1 = –3(2)4 = –3(16) = –48


tenth term: A(10) = –3(2)10 – 1 = –3(2)9 = –3(512) = –1536




                                    8-6
Example 4, page 426

  Find the first, sixth, and twelfth terms of each sequence.
     A) A(n) = 4 • 3n-1            B) A(n) = -2 • 5n-1
             Suppose you drop a tennis ball from a height of 2 meters.
    On each bounce, the ball reaches a height that is 75% of its previous
    height. Write a rule for the height the ball reaches on each bounce. In
    centimeters, what height will the ball reach on its third bounce?

The first term is 2 meters, which is 200 cm.

Draw a diagram to help understand the problem.




                                  8-6
              (continued)
The ball drops from an initial height, for which there is no bounce. The
initial height is 200 cm, when n = 1. The third bounce is n = 4. The
common ratio is 75%, or 0.75.
A rule for the sequence is A(n) = 200 • 0.75n – 1.
                                      Use the sequence to find the height of
         A(n) = 200 • 0.75n – 1
                                      the third bounce.
                                      Substitute 4 for n to find the height of
         A(4) = 200 • 0.754 – 1
                                      the third bounce.
               = 200 • 0.753          Simplify exponents.
               = 200 • 0.421875       Evaluate powers.
               = 84.375               Simplify.

The height of the third bounce is 84.375 cm.


                                   8-6
REAL-WORLD CONNECTION
Example 5, page 426
    You drop a basketball from a height of 2 meters. Each curved
     path has 56% of the height of the previous path. Using the height
     in centimeters, write a rule for the sequence. What height will the
     basketball reach at the top of the fourth path (where n = 4)?
     Round to the nearest tenth of a centimeter.
Summary

    What did you learn today?
SUMMARY

  Look for a pattern.
  Find the common difference (arithmetic)
   or ratio (geometric).
  Substitute for n to find a specific term
   when given a formula for a sequence.
ASSIGNMENT

     #8-6, page 427

       1-43 odd
       51-71 odd

				
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