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Math Help - Geometric series #2

  1. #1
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    Geometric series #2

    Another one is this one that I am unsure how to go about it:

    Find the nth term of the geometric series: 1024, 512, 256, 128,...
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  2. #2
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    Quote Originally Posted by mike1 View Post
    Another one is this one that I am unsure how to go about it:

    Find the nth term of the geometric series: 1024, 512, 256, 128,...
    an = a1 *r^(n-1) --------------***

    1024, 512, 256, 128,...
    So, a1 = 1024.

    r = 512/1024 = 1/2
    r = 256/512 = 1/2
    r = 128/256 = 1/2

    Hence, nth term is
    an = a1 *r^(n-1)
    an = 1024 *(1/2)^(n-1)
    an = 1024 *[(1/2)^n / (1/2)]
    an = 1024 *[2 *(1/2)^n]
    an = 2048 *[1^n / 2^n]
    an = 2048 *[1 / 2^n]
    an = 2048 / 2^n ----------------answer.

    Check,
    a3 = 256 --------given
    256 =? 2048 / (2^3)
    256 =? 2048 / 8
    256 =? 256
    Yes, so, OK.
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  3. #3
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    Hello, Mike!

    Find the nth term of the geometric series: 1024, 512, 256, 128, ...
    If you have the formula and can recognize the parts, you're all set.

    The nth term of a geometric series: a_n .= .ar^{n-1}


    We can see that the first term is: a = 1024, right?

    And we should be able figure out that: r = 1/2

    Then we plug them into the formula . . .


    So where exactly is your difficulty?

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  4. #4
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    My problem is that I am doubting myself. It turns out easier than I make it out to be. You are a great help.
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