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Math Help - Geometric Progression

  1. #1
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    Geometric Progression

    A geometric progression has first term and common ratio both equal to a, a>1. Given that the sum of the first 12 terms is 28 times the sum of the first 6 terms, find the exact value of a.

    S_{12}=28S_6

    \frac{a(1-r^{12})}{1-r}=28\frac{a(1-r^6)}{1-r}

    1-r^{12}=28(1-r^6)

    r^{12}-28r^6+27=0

    I would have to plot a graph to find the answer, but this would not give me an exact answer. How could I solve it in a way which would give me an exact answer?
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  2. #2
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    Quote Originally Posted by Punch View Post
    A geometric progression has first term and common ratio both equal to a, a>1. Given that the sum of the first 12 terms is 28 times the sum of the first 6 terms, find the exact value of a.

    S_{12}=28S_6

    \frac{a(1-r^{12})}{1-r}=28\frac{a(1-r^6)}{1-r}

    1-r^{12}=28(1-r^6)

    r^{12}-28r^6+27=0

    I would have to plot a graph to find the answer, but this would not give me an exact answer. How could I solve it in a way which would give me an exact answer?
    The equation is quadratic in form and can we factored as follows

     (r^6)^2-28r^6+27=0 \iff (r^6-1)(r^6-27)=0

    Note that

    r^6=(r^2)^3

    Use the difference of cubes...

    Can you finish from here?
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    The equation is quadratic in form and can we factored as follows

     (r^6)^2-28r^6+27=0 \iff (r^6-1)(r^6-27)=0

    Note that

    r^6=(r^2)^3

    Use the difference of cubes...

    Can you finish from here?
    could there be another way to complete this? because difference of cubes is something i have not learnt
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  4. #4
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    Quote Originally Posted by Punch View Post
    could there be another way to complete this? because difference of cubes is something i have not learnt
    Then you can consider it something new you're learning now.

    Sum of Two Cubes: \displaystyle a^3 + b^3 = (a + b)(a^2 - ab + b^2).

    Difference of Two Cubes: \displaystyle a^3 - b^3 = (a - b)(a^2 + ab + b^2).
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  5. #5
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    okay, i have solved the first part of the question and now comes another part which i can't solve:

    Hence evaluate log_{3}(\frac{3}{2}a^2+a^4+a^6+....+a^{58}), giving your answer in the form A-log_3B, where A and B are positive integers to be determined.
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