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Math Help - Geometric Sequences

  1. #1
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    Geometric Sequences

    I thank all of you again for helping me, and having the patience to withstand my idiocy. I understand the concept, but maybe it's the wording of the question that is throwing me off, so I am wondering again if any of you could help me out with another question. Thank you all!!

    Directions: Write the first five terms of the geometric sequence. Find the common ratio and write the nth term of the sequence as a function of n.

    a1=64, a^k+1 =1/2a^k

    Edit: Okay I looked at the question, and with the help of my sister, I figured it out. So I guess I can leave the question up just in case anyone is interested in doing it, however I stress again that it is very kind of all of you who have been able to help me. Also I'll leave this thread up in case I have questions with some of the other problems, or for others who are struggling as well in this portion of Precalculus. Thanks!
    Last edited by Norman Smith; September 10th 2007 at 04:07 PM. Reason: I figured the problem out.
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  2. #2
    MHF Contributor
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    a1=64, a^k+1 =1/2a^k

    I guess that is
    a1=64, a^(k+1) = (1/2)a^k

    If I guess it right, then,
    a^(k+1) = (1/2)a^k
    (a^k)(a) = (1/2)a^k
    a = 1/2

    What now?
    a1 = 64 and a = 1/2
    What is this "a"? So what if it is equal to 1/2?
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  3. #3
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    Hello, Norman!

    I believe those "exponents" are actually subscripts.


    Write the first five terms of the geometric sequence.
    Find the common ratio and write the n^{th} term of the sequence as a function of n.

    . . a_1\:=\:64,\;a_{k+1} \:=\:\frac{1}{2}a_k

    We are told that the first term is 64,
    . . and each term is one-half of the preceding term.

    The common ratio is: . r \,=\,\frac{1}{2}

    The first five terms are: . 64,\,32,\,16,\,8,\,4

    The general term is: . a_n \;=\;64\cdot\left(\frac{1}{2}\right)^{n-1} \;=\;\frac{128}{2^n}

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