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Math Help - geometric and arithmetic sequence problem

  1. #1
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    geometric and arithmetic sequence problem

    An arithmetic series has a first term of -4 and a common difference of 1. A geometric series has a first term of 8 and a common ratio of 0.5. After how many terms does the sum of the arithmetic series exceed the sum of the geometric series?

    I end up getting:

    \frac{n}{2}(n-9)>4(1-0.5^n)

    and I'm not sure how to solve it!

    The answer says n=12.
    Last edited by shawli; April 5th 2010 at 11:22 AM.
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  2. #2
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    So the arithmetic series is -4-3-2-1+0+1+2+... and the geometric series is 8+ 4+ 2+ 1+ \frac{1}{2}+ \frac{1}{4}+ ....

    The nth term in the arithmetic series is -4+ (n-1) and the sum up to that n is (-4+ -4+ n-1)n/2= n(n- 9)/2, exactly what you have.

    The sum of the geometric series is \frac{8(1- 0.5^{n-1})}{1- 0.5}= 16(1- 0.5^{n-1}) which is almost what you have. Looks like you multiplied by 1/2 rather than dividing.

    Now you want n(n-9)/2> 16(1- 0.5^{n-1}
    Notice that the infinite serie would have sum 16. n(n-9)/2> 16 when n(n- 9)= n^2- 9n> 32 or n^2- 9n- 32> 0.

    n^2- 9n- 32= 0 when n= \frac{9\pm\sqrt{81+ 128}}{2}= \frac{9\pm\sqrt{209}}{2} which is about 11.7.
    Now just check: when n= 11, the arithmetic sum is 11(11-9)/2= 11 and the geometric sum is 16(1- .5^10)= 15.984375. When n= 12, the arithmetic sum is 12(12- 9)/2= 18 and the geometric sum is [tex]16(1- .5^{11})= 15.9921875.
    Last edited by HallsofIvy; April 6th 2010 at 04:28 AM.
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  3. #3
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    Ohhhh I was supposed to use the fact that it was an infinite series there (and also not mess up my algebra).

    Thank you!
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