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Math Help - geometric sequence

  1. #1
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    geometric sequence

    How many terms are in each sequence?

    12, 4, 4/3, ......, 4/729


    help i dont know how to get answer
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  2. #2
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    Re: geometric sequence

    Your terms are:

    12, \frac{12}{3}, \frac{12}{9}, \frac{12}{27}......\frac{12}{2187}

    Or

    \frac{12}{3^0}, \frac{12}{3^1}, \frac{12}{3^2}, \frac{12}{3^3}......\frac{12}{3^7}
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  3. #3
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    Re: geometric sequence

    Hello, Vince604!

    How many terms are in the sequence?

    . . 12,\;4,\;\tfrac{4}{3}\;\hdots\;\tfrac{4}{729}

    Help, I don't know how to get answer. . You don't?
    If you can't see the pattern, you need more help than we can offer.


    The terms are successively divided by three.

    So, the very least you can do is list the terms.

    . . \begin{array}{ccccc}\text{start} && 12 & (1) \\ 12\div 3 &=& 4 & (2) \\ 4\div 3 &=& \frac{4}{3} & (3) \\ \\[-4mm] \frac{4}{3} \div 3 &=& \frac{4}{9} & (4) \\ \\[-4mm] \frac{4}{9} \div 3 &=& \frac{4}{27} & (5) \\ \\[-4mm] \frac{4}{27} \div 3 &=& \frac{4}{81} & (6) \\ \\[-4mm] \frac{4}{81}\div3 &=& \frac{4}{243} & (7) \\ \\[-4mm] \frac{4}{243}\div3 &=& \frac{4}{729} & (8) \end{array}

    Gee, it looks like eight terms.
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  4. #4
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    Re: geometric sequence

    Quote Originally Posted by Soroban View Post
    Hello, Vince604!

    If you can't see the pattern, you need more help than we can offer.


    The terms are successively divided by three.

    So, the very least you can do is list the terms.

    . . \begin{array}{ccccc}\text{start} && 12 & (1) \\ 12\div 3 &=& 4 & (2) \\ 4\div 3 &=& \frac{4}{3} & (3) \\ \\[-4mm] \frac{4}{3} \div 3 &=& \frac{4}{9} & (4) \\ \\[-4mm] \frac{4}{9} \div 3 &=& \frac{4}{27} & (5) \\ \\[-4mm] \frac{4}{27} \div 3 &=& \frac{4}{81} & (6) \\ \\[-4mm] \frac{4}{81}\div3 &=& \frac{4}{243} & (7) \\ \\[-4mm] \frac{4}{243}\div3 &=& \frac{4}{729} & (8) \end{array}

    Gee, it looks like eight terms.
    Is there a faster way (like a formula i can use) to find the eighth term without having to list all the terms to get to it? Thanks for the help btw
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  5. #5
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    Re: geometric sequence

    Hello again, Vince604!

    Is there a faster way (like a formula i can use) to find the eighth term,
    without having to list all the terms to get to it?

    Yes, there is a formula.

    We have a geometric series.
    The first term is a = 12; the common ratio is r = \tfrac {1}{3}

    The nth term is: . a_n \:=\:ar^{n-1} \:=\:12(\tfrac{1}{3})^{n-1}

    When does: . 12\cdot\frac{1}{3^{n-1}} \:=\:\frac{4}{729} ?

    Solve for n.
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  6. #6
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    Re: geometric sequence

    Quote Originally Posted by Soroban View Post
    Hello again, Vince604!


    Yes, there is a formula.

    We have a geometric series.
    The first term is a = 12; the common ratio is r = \tfrac {1}{3}

    The nth term is: . a_n \:=\:ar^{n-1} \:=\:12(\tfrac{1}{3})^{n-1}

    When does: . 12\cdot\frac{1}{3^{n-1}} \:=\:\frac{4}{729} ?

    Solve for n.
    umm don't know
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