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Math Help - A Level Maths part2

  1. #1
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    Talking A Level Maths part2

    really having touble with all that need urgent help!!!

    regards,
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by carlasader View Post
    really having touble with all that need urgent help!!!

    regards,
    1)
    a geometric series is one of the form: \sum_{n=1}^{\infty} ar^{n-1}

    where a is the first term, r is the common ratio and n is the current number of the term.

    a)
    we are given that the second term is 80, \Rightarrow ar = 80
    we are also told that the fifth term is 5.12, \Rightarrow ar^4 = 5.12

    \Rightarrow \frac {ar^4}{ar} = \frac {5.12}{80}

    \Rightarrow r^3 = 0.064

    \Rightarrow r = 0.4 \mbox{ or } \frac {4}{10}

    since ar = 80 \mbox{ we have } a = \frac {80}{r} = \frac {80}{ \frac {4}{10}} = 200

    so the first term is 200, and the common ration is \frac {4}{10}


    b)
    For a geometric series with \left| r \right| < 1, the sum to infinity is given by: S_{ \infty} = \frac {a}{1 - r}

    So here, S_{ \infty} = \frac {200}{1 - \frac {4}{10}} = 333 \frac {1}{3}

    c)
    The sum of the first n terms of a geometric series is given by S_{n} = \frac {a(1 - r^n)}{1 - r}
    So S_{ \infty} - S_{14} = 333 \frac {1}{3} - \frac {200(1 - 0.4^{14})}{1 - 0.4} = 0.0008947 = 9 \times 10^{-4}
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  3. #3
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    Talking thanks

    can u please help me with the rest of the homework...anyone
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    i believe the other questions were answered in your subsequent posts
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