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Math Help - Geometric progression

  1. #1
    Newbie parmis's Avatar
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    Geometric progression

    in a geometric progression a4 * a5=1/64
    a1*a2*a3*a4....*a13=?
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  2. #2
    Newbie parmis's Avatar
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    Re: Geometric progression

    and these question's answer is 40?
    In a Geometric progression :
    a5 is 24 more than a3. => a5-a3=24
    if terms are Integer and positive. a3+a5=?
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  3. #3
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    Re: Geometric progression

    Quote Originally Posted by parmis View Post
    in a geometric progression a4 * a5=1/64
    a1*a2*a3*a4....*a13=?
    A geometric progression is one of the form a, ar, ar^2, ar^3,\ldots. So, suppose a_0 = a, a_1 = ar, a_2 = ar^2, a_3 = ar^3, a_4 = ar^4, a_5 = ar^5. Then a_4 a_5 = (ar^4)(ar^5) = a^2r^9 = \dfrac{1}{64}. So, \prod_{n=1}^{13} a_n = a^{13}r^{91} = (a^2r^9)^6(ar^{37}) = \left(\dfrac{1}{64}\right)^6(ar^{37}). You do not have enough information to solve this.

    Edit: We can actually do a little better. Since a\neq 0, we can solve for r. a^2r^9 = \dfrac{1}{64} \Rightarrow r^9 = (8a)^{-2} \Rightarrow r = (8a)^{-2/9}. Then \prod_{n=1}^{13} a_n = a^{13}r^{91} = a^{13}(8a)^{-182/9} = a^{-65/9}2^{-182/3}. That's the closest I can come to a solution.
    Last edited by SlipEternal; October 19th 2013 at 09:23 AM.
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  4. #4
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    Re: Geometric progression

    Quote Originally Posted by parmis View Post
    and these question's answer is 40?
    In a Geometric progression :
    a5 is 24 more than a3. => a5-a3=24
    if terms are Integer and positive. a3+a5=?

    Terms are integers. So, a_3 = ar^3, a_5 = ar^5. Suppose a=1. Then a_5-a_3 = r^5-r^3=24 has r=2 as a solution. For any integer a>1, there are no integer solutions to a(r^5-r^3) = 24. So, that must be the correct solution. Then a_3+a_5 = r^3+r^5 = 2^3+2^5 = 8 + 32 = 40.
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  5. #5
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    Re: Geometric progression

    Hello, parmis!

    There are at least 4 possible solutions . . .



    \text{In a geometric progression: }\;a_4\cdot a_5\,=\,\tfrac{1}{64}

    \text{Find: }\:P \;=\;a_1\cdot a_2\cdot a_3 \cdots a_{13}

    We are given: . a_4\cdot a_5 \,=\,\tfrac{1}{64}

    I found four cases: . (a_4,a_5) \:=\:\left(\tfrac{1}{16},\tfrac{1}{4}\right),\; \left(\tfrac{1}{4},\tfrac{1}{16}\right),\;\left( \text{-}\tfrac{1}{16},\text{-}\tfrac{1}{4}\right),\;\left(\text{-}\tfrac{1}{4},\text{-}\tfrac{1}{64}\right)


    Take the first case: . \begin{Bmatrix}a_4 &=& \frac{1}{16} \\ a_5 &=& \frac{1}{4} \end{Bmatrix}

    We find that: . a = \tfrac{1}{4^5},\;r = 4


    Then: . P \;=\;a_1\cdot a_2\cdot a_3\cdot a_4 \cdots a_{13}

    . . . . . . . . =\; (a)(ar)(ar^2)(ar^3) \cdots (ar^{12})

    . . . . . . . . =\; a^{13}r^{78}


    . . . . . P \;=\;\left(\tfrac{1}{4^5}\right)^{13}(4)^{78} \;=\;\tfrac{1}{4^{65}}\cdot4^{78}

    . . . . . . . =\;4^{13} \;=\;67,\!108,\!864


    You can work out the other three solutions.
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