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Math Help - geometric progression

  1. #1
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    geometric progression

    A geometric progression has first term a where a‡0 and common ratio r where r‡1. the difference between the fourth and first term is equal to four times the difference between the third term and the second term.

    i) show that r - 4r + 4r -1 = 0
    ii) find two possible values for the ratio of the geometric progression.
    ii) for the value of r which the progression is convergent, prove that the sum to infinity is 1/2 a (1 + √5)

    thanks
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  2. #2
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    The nth term in a geometric sequence is given by

    a_{n}=a_{1}r^{n-1}


    For part a you have a_{1}r^{3}-a_{1}=4(a_{1}r^{2}-a_{1}r)

    a_{1}r^{3}-a_{1}=4a_{1}r^{2}-4a_{1}r

    a_{1}r^{3}-4a_{1}r^{2}+4a_{1}r-a_{1}

    Factor out a_{1}:

    a_{1}(r^{3}-4r^{2}+4r-1)

    Solve the cubic to find the possible values of r.
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  3. #3
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    Hello, Gracey!

    A geometric progression has first term a \neq 0 and common ratio r \neq 1
    The difference between a_4\text{ and }a_1 is equal to 4 times the difference between a_3\text{ and }a_2

    (a) Show that: . r^3 - 4r^2 + 4r - 1 \:=\:0
    The first four terms are: . \begin{Bmatrix}a_1 &=& a \\ a_2 &=& ar \\ a_3 &=& ar^2 \\ a_4 &=&ar^3 \end{Bmatrix}


    We are told that: . a_4 - a_1 \:=\:4\left(a_3-a_2\right)

    So we have . ar^3 - a \:=\:4(ar^2-ar) \quad\Rightarrow\quad ar^3 - 4ar^2 + 4ar - a \:=\:0

    . . Divide by a\!:\;\;\boxed{r^3 - 4r^2 + 4r - 1 \:=\:0}




    (b) Find two possible values for r.
    Factor: . (r-1)(r^2-3r+1) \:=\:0


    We have: . r-1\:=\:\quad\Rightarrow\quad r \,=\,1\quad\hdots \text{ which is not allowed}

    . . and: . r^2-3r+1\:=\:0\quad\Rightarrow\quad \boxed{r \:=\:\frac{3\pm\sqrt{5}}{2}}




    (c) For the value of r which the progression is convergent,
    prove that the sum to infinity is: . \frac{1}{2}a\left(1 + \sqrt{5}\right)
    To be convergent,  r < 1. . Hence, we will use: . r \;=\;\frac{3-\sqrt{5}}{2}

    Formula: . S \;=\;\sum^{\infty}_{n=0} ar^n \;=\;\frac{a}{1-r}

    \text{So we have: }\;S \;=\;\frac{a}{1 - \frac{3-\sqrt{5}}{2}} \;=\;\frac{2a}{\sqrt{5}-1}

    Rationalize: . S \;=\;\frac{2a}{\sqrt{5}-1}\cdot\frac{\sqrt{5}+1}{\sqrt{5}+1} \;=\;\frac{2a(\sqrt{5}+1)}{5-1}\;=\;\frac{2a(1 + \sqrt{5})}{4}

    . . Therefore: . \boxed{S \;=\;\frac{1}{2}a(1 + \sqrt{5})}

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