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Thread: Sequence

  1. #1
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    Sequence

    The 2nd, 6th and 8th terms of an Arithmetic progression are three successive terms of a Geometric.P.Find the common ratio of G.P and obtain an expression for nth term of the G.P. (question)
    my first step is to find the common ratio, and i get r=2.....but i cant obtain others constant to find the expression...anyone know how to find that?


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  2. #2
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    Re: Sequence

    Quote Originally Posted by goldsomecriss View Post
    my first step is to find the common ratio, and i get r=2
    HOW did you get that?
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    Re: Sequence

    Set the starting point in the arithmetic sequence asa2=a1+k

    Then we have:

    a2=a1+k=br0 ..................Equation(1)
    a6=a1+5k=br1 ................Equation(2)
    a8=a1+7k=br2 ................Equation(3)

    Eqiation(2)−Equation(1) → 4k=br−b

    → 4k=b(r−1)..Eqn(4)

    Equation(3)−Equation(2) → 2k=br2−br

    → 2k=br(r−1)..Eqn(5)

    Eqn(5)Eqn(4) → 4k2k=br(r−1)b(r−1)

    → 2.= r



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    Re: Sequence

    Quote Originally Posted by DenisB View Post
    HOW did you get that?
    do u know the next step?

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  5. #5
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    Re: Sequence

    Arithmetic: 1st term = 1, common difference = 0
    1,1,1,1,1,1,1,1,1,.........

    Geometric: 1st term = 1, common ratio = 1
    1,1,1,1,1,1,1,1,1,.........

    YES, I'm lazy
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  6. #6
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    Re: Sequence

    Quote Originally Posted by DenisB View Post
    Arithmetic: 1st term = 1, common difference = 0
    1,1,1,1,1,1,1,1,1,.........

    Geometric: 1st term = 1, common ratio = 1
    1,1,1,1,1,1,1,1,1,.........

    YES, I'm lazy
    Lol, all zeros would work, as well. But, the common ratio may be any real number.
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  7. #7
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    Re: Sequence

    The 2nd, 6th and 8th terms of an Arithmetic progression are three successive terms of a Geometric.P.Find the common ratio of G.P and obtain an expression for nth term of the G.P. (question)
    $a_2 = a_1 + d = b_1$

    $a_6 = a_1 + 5d = b_1 \cdot r$

    $a_8 = a_1 + 7d = b_1 \cdot r^2$

    from the first two equations, $d = \dfrac{b_1(r-1)}{4}$

    from the first and third equation $d = \dfrac{b_1(r^2-1)}{6}$


    $\dfrac{b_1(r-1)}{4} = \dfrac{b_1(r-1)(r+1)}{6}$

    for $b_1 \ne 0$ and $r \ne 1$ ...

    $\dfrac{1}{4} = \dfrac{r+1}{6} \implies r = \dfrac{1}{2}$

    $b_n = b_1 \cdot \left(\dfrac{1}{2}\right)^{n-1}$
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