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Math Help - Ap/gp

  1. #1
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    Ap/gp

    The question goes:

    The first, second and fourth terms of a convergent geometric progression are consecutive terms of an arithmetic progression. Prove that the common ratio of the geometric progression is (-1 + sq root 5)/2.

    The first term of the geometric progression is positive. Show that the sum of the first 5 terms of this progression is greater than the nine tenths of the sum to infinity.

    I tried but I couldnt get the first part. As for the second part, I don't really understand the 'nine tenths' part?

    Thanks in advance if you could help!
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  2. #2
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    Hello, margaritas!

    Here's the first part . . .


    The first, second and fourth terms of a convergent geometric progression
    are consecutive terms of an arithmetic progression.
    Prove that the common ratio of the geometric progression is \frac{-1 + \sqrt{5}}{2}

    The first, second and fourth terms of a GP are: a,\:ar,\:ar^3

    Three consecutive terms of an AP are: a,\:a+d,\:a+2d

    [Yes, I let a be the first term of both progressions.]

    Then we have: . \begin{array}{cc}ar = a+ d\quad\Rightarrow\quad ar - a = d\quad\Rightarrow\quad a(r - 1) = d \\ ar^3 = a + 2d\quad\Rightarrow\quad ar^3 - a = d\quad\Rightarrow\quad a(r^3 - 1) = d\end{array}\,\begin{array}{cc}(1)\\(2)\end{array}

    Divide (2) by (1): . \frac{a(r^3 - 1)}{a(r - 1)} \:=\: \frac{2d}{d}\quad\Rightarrow\quad \frac{a(r-1)(r^2+r+1)}{a(r-1)}\:=\:\frac{2d}{d}

    And we have: . r^2 + r + 1 \:=\:2\quad\Rightarrow\quad r^2 + r - 1\:= \:0

    The Quadratic Formula gives us: . r\:=\:\frac{-1\pm\sqrt{1^2 - 4(1)(-1)}}{2(1)}\;=\;\frac{-1\pm\sqrt{5}}{2}

    Therefore, the ratio is: . \boxed{r \:= \:\frac{-1+\sqrt{5}}{2}}


    Note: We are told that the geometric series is convergent.
    . . . . .Hence: |r| < 1
    . . . . .We must discard the other quadratic root.

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  3. #3
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    Oh now I get it, thanks so much for your help Soroban!
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