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Thread: 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

    $\displaystyle a_{n}=a_{1}r^{n-1}$


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

    $\displaystyle a_{1}r^{3}-a_{1}=4a_{1}r^{2}-4a_{1}r$

    $\displaystyle a_{1}r^{3}-4a_{1}r^{2}+4a_{1}r-a_{1}$

    Factor out $\displaystyle a_{1}$:

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

    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 $\displaystyle a \neq 0$ and common ratio $\displaystyle r \neq 1$
    The difference between $\displaystyle a_4\text{ and }a_1$ is equal to 4 times the difference between $\displaystyle a_3\text{ and }a_2$

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


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

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

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




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


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

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




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

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

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

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

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

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