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Math Help - geometric sequence help

  1. #1
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    geometric sequence help

    Find the tenth term of a geometric sequence.

    the first term is 3 the 6th term is root3/9

    i stuck on the calucation for finding R

    tn=ar^n-1
    t6= root3/9 = 3(r)^5

    (3^-1/2)/3^2=3(r)^5 <-- stuck

    I checked the answer on my workbook .It says.

    (3^-1/2)/3^3=r^5
    3^-5=r^5

    then it suddenly jumps right into r= 3^-1/2

    i do not understand how they get rid of the exponent 5 in the workbook and if i factor out the 9 to 3^2, then how can i continue? will the answer be equivalent?
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  2. #2
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    Using the information given

    \frac{\sqrt{3}}{9}= 3r^5

    Divide by 3 and take the 5th root.
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  3. #3
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    Quote Originally Posted by hovermet View Post
    Find the tenth term of a geometric sequence.

    the first term is 3 the 6th term is root3/9

    i stuck on the calucation for finding R

    tn=ar^n-1
    t6= root3/9 = 3(r)^5

    (3^-1/2)/3^2=3(r)^5 <-- stuck

    I checked the answer on my workbook .It says.

    (3^-1/2)/3^3=r^5
    3^-5=r^5

    then it suddenly jumps right into r= 3^-1/2

    i do not understand how they get rid of the exponent 5 in the workbook and if i factor out the 9 to 3^2, then how can i continue? will the answer be equivalent?
    I am having trouble following your notation. Please use parentheses to clear up confusion.

    If you mean \frac{\sqrt{3}}{9}=3r^5 then cancel out a 3 from both sides, then take the 5th root of both sides.
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