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Geometric progression

  1. Aug 11, 2011 #1
    1. The problem statement, all variables and given/known data

    3,6,12........1536
    determine the number of terms in the progression

    2. Relevant equations



    3. The attempt at a solution
    a=3 r=2

    n= ar^n-1

    1536= (3) (2)^n-1
     
  2. jcsd
  3. Aug 11, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Okay, so solve it! First, divide both sides by 3. If that sequence is actually geometric, you should be able to identify 1536/3 as a power of 2. I suggest you just calculate powers of 2: 1, 2, 4, 8, 16, ... until you get to that number.
     
  4. Aug 11, 2011 #3
    1536= (3) (2)^n-1

    1536/3 = 6^n-1[itex]/[/itex]3

    512 = 2^n-1

    512 = 2^10-1

    512 = 2^9

    n = 10
     
  5. Aug 11, 2011 #4

    Mark44

    Staff: Mentor

    There are errors in your work. Also, you need parentheses around your exponent expressions.
    The above is incorrect. 3*2^(n - 1) [itex]\neq[/itex] 6^(n - 1)
    The above is also incorrect. [6^(n - 1)]/3 [itex]\neq[/itex] 2^(n - 1)
     
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