Results 1 to 7 of 7

Math Help - Arithmetic Series

  1. #1
    Junior Member
    Joined
    Nov 2009
    Posts
    42

    Arithmetic Series

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

    I know that S_{n} = \frac{1}{2}\left[{2a+(n-1)d}\right], but I somehow can't apply it to this problem.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Quote Originally Posted by Captcha View Post
    I'm trying to find the number of terms in the arithmetic series -4, 2, 8, 14, .... which makes the sum strictly greater than 18,000.

    I know that 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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2009
    Posts
    42
    Quote Originally Posted by Jameson View Post
    You know a_1 and d. Set S_n=18,000 and solve for n.
    a = -4

     d = 6

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

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

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

    \frac{1}{2}\left[6n-14\right] = 18000<br />

    3 n-7 = 18,000

    3n = 18,000 -7

    3n = 17993

     n = \frac{17993}{3}

    n = 5997.666667

    5998 terms?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Almost. You subtracted 7 but should have added.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2009
    Posts
    42
    Quote Originally Posted by Jameson View Post
    Almost. You subtracted 7 but should have added.
    Oh, silly mistake.


    3n = 18,000 +7

    3n = 18007

    n = \frac{18007}{3}

    n = 6002.333333

    6003 terms then?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    Seems to make sense.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2009
    Posts
    42
    Many thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. arithmetic series
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: December 8th 2010, 04:07 PM
  2. Arithmetic Series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 9th 2010, 10:56 AM
  3. Arithmetic Progression or Arithmetic Series Problem
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 8th 2009, 12:36 AM
  4. Arithmetic Series Help!?
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 16th 2009, 06:43 AM
  5. arithmetic series
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 18th 2008, 05:36 PM

Search Tags


/mathhelpforum @mathhelpforum