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Thread: Arithmetic Series

  1. #1
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    Arithmetic Series

    I'm trying to find the number of terms in the arithmetic series $\displaystyle -4, 2, 8, 14, ....$ which makes the sum strictly greater than $\displaystyle 18,000$.

    I know that $\displaystyle S_{n} = \frac{1}{2}\left[{2a+(n-1)d}\right]$, but I somehow can't apply it to this problem.
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  2. #2
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    Quote Originally Posted by Captcha View Post
    I'm trying to find the number of terms in the arithmetic series $\displaystyle -4, 2, 8, 14, ....$ which makes the sum strictly greater than $\displaystyle 18,000$.

    I know that $\displaystyle S_{n} = \frac{1}{2}\left[{2a+(n-1)d}\right]$, but I somehow can't apply it to this problem.
    You know a_1 and d. Set S_n=18,000 and solve for n.
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  3. #3
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    Quote Originally Posted by Jameson View Post
    You know a_1 and d. Set S_n=18,000 and solve for n.
    $\displaystyle a = -4 $

    $\displaystyle d = 6$

    $\displaystyle \frac{1}{2}\left[2(-4)+(n-1)6\right] = 18000$

    $\displaystyle \frac{1}{2}\left[-8+6n-6\right] = 18000$

    $\displaystyle \frac{1}{2}\left[-8+6n-6\right] = 18000$

    $\displaystyle \frac{1}{2}\left[6n-14\right] = 18000
    $

    3$\displaystyle n-7 = 18,000$

    $\displaystyle 3n = 18,000 -7$

    $\displaystyle 3n = 17993$

    $\displaystyle n = \frac{17993}{3}$

    $\displaystyle n = 5997.666667$

    $\displaystyle 5998$ terms?
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  4. #4
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    Almost. You subtracted 7 but should have added.
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  5. #5
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    Quote Originally Posted by Jameson View Post
    Almost. You subtracted 7 but should have added.
    Oh, silly mistake.


    $\displaystyle 3n = 18,000 +7 $

    $\displaystyle 3n = 18007$

    $\displaystyle n = \frac{18007}{3}$

    $\displaystyle n = 6002.333333$

    $\displaystyle 6003$ terms then?
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  6. #6
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    Seems to make sense.
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  7. #7
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    Many thanks!
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