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Geometric Sequences

  1. Jul 27, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi, I was trying to work out this question, but i kinda got stuck. Can anyone help me please?
    Thanks

    4. Find the number of terms in each of these geometric sequences.

    2,10,50....1250


    2. Relevant equations

    ar^n-1


    3. The attempt at a solution

    1250=2x5^n-1
     
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  3. Jul 27, 2009 #2

    Cyosis

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    I take it you mean [itex]1250=2*5^{n-1}[/itex] with n starting at 1? You're on the right track. What is the problem with the expression you found so far? You are unable to solve it for n? Hint: logarithm
     
    Last edited: Jul 27, 2009
  4. Jul 27, 2009 #3
    i don't get what the next step is
     
  5. Jul 27, 2009 #4

    Cyosis

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    What does n represent in that formula?
     
  6. Jul 27, 2009 #5
    n is the term, but i dont know how to find it
     
  7. Jul 27, 2009 #6

    Cyosis

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    "The term", you mean is the number of terms in the sequence. The first step in solving it is to put 5^(n-1) on one side and the other terms on the other side. Then take the logarithm on both sides.
     
  8. Jul 27, 2009 #7
    so do i have to balance both sides and eliminate 1250?
     
  9. Jul 27, 2009 #8

    Cyosis

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    An equation is balanced by definition. If you substitute x=5^(n-1) then the equation becomes 1250=2x. Now solve this equation for x and then back substitute x.

    ps. Are you familiar with logarithms?
     
  10. Jul 27, 2009 #9
    Thank you very much. I've just started learning about it, so I'm trying to get the hang of it, but thank you for helping me.
     
  11. Jul 27, 2009 #10

    Cyosis

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    You're welcome, but have you found the solution?
     
  12. Jul 27, 2009 #11
    1250= 2x so x=1250/2 x=625

    oh no, I'm getting confused again. sorry
     
  13. Jul 27, 2009 #12

    Cyosis

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    Correct x=625 and because we substituted x=5^(n-1) we know that 5^(n-1)=625. We now want to write 625 in terms of 5 raised to a certain exponent. For example we can write 8=2^3, 125=5^3. Try to write 625 in a similar way.
     
  14. Jul 27, 2009 #13
    so it can be 625=25^2
     
  15. Jul 27, 2009 #14

    Cyosis

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    Yes that is correct although not entirely what I had in mind. Note that you can write 25=5^2 therefore 25^2=(5^2)^2=5^4=625.

    So now your equation becomes:

    [tex]
    5^{n-1}=5^4
    [/tex]

    Therefore n-1 equals...?
     
  16. Jul 27, 2009 #15
    n-1= 5
     
  17. Jul 27, 2009 #16

    Cyosis

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    No the two exponents need to be equal to each other, it is the only way the equation can hold. So I will ask you again n-1=?

    What you're saying now is that 5^5=5^4. Convince yourself that this cannot be correct.
     
  18. Jul 27, 2009 #17
    ohhhh n-1= 5-1?
     
  19. Jul 27, 2009 #18

    Cyosis

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    Why do you think so, explain your reasoning behind it.
     
  20. Jul 27, 2009 #19
    because 5^(5-1) which equals to 5^4
     
  21. Jul 27, 2009 #20

    Cyosis

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    Well your reasoning and your answer are correct although a bit of a detour. You know that both exponents need to be the same therefore n-1=4.
     
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