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Math Help - Geometric Sequences + Irrational Numbers

  1. #1
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    Geometric Sequences + Irrational Numbers

    Question:
    find the general term of Un which as
    u3=3 and u9=3/8

    --> i am looking for the common ratio atm and this is what i did
    u3=u1r^2=3 (1)
    u9=u1r^(-5/8)=3/8 (2)

    --> (2)/(1)
    u1r^(-5/8)/ u1r^2 =3/8 / 3

    r^(-21/8) = 1/8

    r = (-21/8)* root 1/8

    But i don't know how to simplify it from there..

    The answer said 1/root2
    but i don't know how to get that

    Help please?
    thank you
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  2. #2
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    Quote Originally Posted by juliak View Post
    Question:
    find the general term of Un which as
    u3=3 and u9=3/8

    --> i am looking for the common ratio atm and this is what i did
    u3=u1r^2=3 (1)
    u9=u1r^(-5/8)=3/8 (2)

    --> (2)/(1)
    u1r^(-5/8)/ u1r^2 =3/8 / 3

    r^(-21/8) = 1/8

    r = (-21/8)* root 1/8

    But i don't know how to simplify it from there..

    The answer said 1/root2
    but i don't know how to get that

    Help please?
    thank you
    u_3 = a r^2 and u_9 = a r^8. Therefore:

    \frac{u_9}{u_3} = r^6 = \frac{3/8}{3}\Rightarrow r^6 = \frac{1}{8} \Rightarrow r^6 = \frac{1}{2^3} = ....
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  3. #3
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    u9=a1r^-5/8?

    how does u9=a1r^8?
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  4. #4
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    Quote Originally Posted by juliak View Post
    u9=a1r^-5/8?

    how does u9=a1r^8?
    Didn't you say it's a geometric series.

    You're expected to know that the nth term is given by u_n = a r^{n-1}.

    I have absolutely no idea where you're getting u9=a1r^-5/8 from.
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  5. #5
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    oh >_<

    but umm i stil don't really get how to simplify
    r^6 = 1/2^3
    why do you have to change 8 to 2^3 and what do you do to continue simplifying it?
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  6. #6
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    Quote Originally Posted by juliak View Post
    oh >_<

    but umm i stil don't really get how to simplify
    r^6 = 1/2^3
    why do you have to change 8 to 2^3 and what do you do to continue simplifying it?
    You should go back and review all the material (like index laws) that this question clearly implies you should already know.

    r^6 = \frac{1}{2^3} \Rightarrow r = \left( \frac{1}{2^3} \right)^{1/6} = \frac{1}{(2^3)^{1/6}} = \frac{1}{2^{1/2}} = \frac{1}{\sqrt{2}}.
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