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Geometric Series

  1. Sep 6, 2010 #1
    In the geometric sequence above, b1=1000 and bn=(2/3)bn-1 for all n[tex]\geq[/tex]2. What is the least value of k for which bk<0.001?

    3. The attempt at a solution
    What I did first was I found what b0 is since we are given b1 and that is 1500. But I do not understand where the k is coming from and what needs to be found. Can someone please help me?
    Thank you!
  2. jcsd
  3. Sep 6, 2010 #2


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    There's no need to find b0. b1=1000. b2=1000*(2/3). That's bk for k=1. b3=1000*(2/3)*(2/3), right? That's bk for k=3. None of those is less that 0.001. But they are getting smaller. How large does k have to be to make bk<0.001?
  4. Sep 6, 2010 #3


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    Its asking you for which bk is the sequence starting to be less than 0.001

    According to you.. b0 = 1500, b1= 1000, b2=(2/3)*1000, b3=(2/3)*b2, b4=(2/3)*b3, ... bk=(2/3)*bk-1

    0.001 is 1/(1000). So how many times would you have to multiply 1500*(2/3)^x to equal 1/1000 ?

    1500*(2/3)^x=1/1000. Find x, and your answer would be for k>x

    Edit: I see Dick beat me to it
    Last edited: Sep 6, 2010
  5. Sep 6, 2010 #4
    Thank you both for your help!

    Now I understand what the problem is asking. So if I continue to plug in each bk i get and try to find the one where bk<.001 the answer would be 36?
  6. Sep 6, 2010 #5


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