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Thread: Series and Sequences Part Two

  1. January 9th 2010, 03:33 AM #1
    christina
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    Series and Sequences Part Two

    Insert 4 terms in geometric sequence between 3/8 and 128/81

    Find a number which when added to each of -1,1 and 5 gives 3 terms in geometric sequence

    The sum of the first 8 terms of a geometric series is 17 times the sum of its first 4 terms. Find the common ratio.

    Thank you for any help given on these questions !
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  2. January 9th 2010, 07:50 AM #2
    Grandad
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    Hello christina
    Quote Originally Posted by christina View Post
    Insert 4 terms in geometric sequence between 3/8 and 128/81
    The first term a = \frac38. The sixth term = ar^5 = \frac{128}{81}

    \Rightarrow r^5 = \frac{128}{81}\cdot\frac{8}{3}

    This will give you the value of r, and then you can work out the terms in between.
    Spoiler:
    r = \frac{4}{3}; the terms in between are \frac{1}{2},\frac{2}{3},\frac{8}{9}, \frac{32}{27}

    Find a number which when added to each of -1,1 and 5 gives 3 terms in geometric sequence
    Let the number be x. Then the terms (x-1),(x+1) and (x+5) are in geometric sequence. In other words:
    \frac{x-1}{x+1}=\frac{x+1}{x+5}
    Solve this equation for x.
    Spoiler:
    x=3

    The sum of the first 8 terms of a geometric series is 17 times the sum of its first 4 terms. Find the common ratio.
    With the usual notation, the sum of the first n terms is:
    \frac{a(r^n-1)}{r-1}
    So:
    \frac{a(r^8-1)}{r-1}=\frac{17a(r^4-1)}{r-1}

    \Rightarrow r^8-1=17(r^4-1)
    This is easy to solve if you spot that r^8-1=(r^4-1)(r^4+1), and so the equation simplifies considerably!
    Spoiler:
    r^4+1 = 17 \Rightarrow r = \pm2


    Grandad
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  3. January 9th 2010, 06:57 PM #3
    christina
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    thank you so much Grandad!! I really appreciate your help, thank you for taking the time to answer my questions ! =)
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