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    Sequence

    Find the third element of a geometric sequence who fifth element is 81 and whose ninth element is 16. Help please!
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  2. #2
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    Quote Originally Posted by reiward View Post
    Find the third element of a geometric sequence who fifth element is 81 and whose ninth element is 16. Help please!
    Hi reiward,

    You should have been given the formula for finding the nth term of a geometric sequence.

    a_n=a_1 \cdot r^{n-1}

    r = common ratio

    Your first element ( a_1) will be 81.

    n = 5 because there are five elements in all between the 5th element and the 9th element, inclusive.

    a_n = 16

    Solve for r

    Now that you have found the common ratio, substitute back into the original formula and solve for a_3.

    This time there are 3 elements from the 3rd to the 5th elements, inclusive. So, n=3

    Remember a_5=81

    a_5=a_3 \cdot r^{3-1}
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  3. #3
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    Hello, reiward!

    Use the formula for the n^{th} term: . a_n \:=\:ar^{n-1}


    Find the third element of a geometric sequence who fifth element is 81
    and whose ninth element is 16.
    We are given: . \begin{array}{ccccccc}a_5 &=& ar^4 &=& 81 & [1] \\ a_9 &=& ar^8 &=& 16 & [2] \end{array}

    Divide [2] by [1]: . \frac{ar^8}{ar^4} \:=\:\frac{16}{81} \quad\Rightarrow\quad r^4 \:=\:\frac{16}{81} \quad\Rightarrow\quad r \:=\:\pm\frac{2}{3}

    Substitute into [1]: . a\left(\pm\frac{2}{3}\right)^4 \:=\:81 \quad\Rightarrow\quad \frac{16}{81}a\:=\:81 \quad\Rightarrow\quad a \:=\:\frac{6561}{16}


    Therefore: . a_3\;=\;ar^2 \;=\;\frac{6561}{16}\left(\pm\frac{2}{3}\right)^2 \;=\;\frac{729}{4}

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