Results 1 to 6 of 6

Math Help - Arithmetic progression

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    202

    Arithmetic progression

    The sum of the first term and seventh term of an arithmetic progression is 6.If the twentieth term is 56,find the term whose value first exceeds 400.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Jun 2009
    Posts
    220
    Thanks
    1
    Quote Originally Posted by mastermin346 View Post
    The sum of the first term and seventh term of an arithmetic progression is 6.If the twentieth term is 56,find the term whose value first exceeds 400.
    The terms are in arithmetic progression, so lets call the difference between terms d, and the first term a. Then we are given that the first term and seventh term sum to 6, therefore,
    a + (a + 6d) = 6 .
    We are then given that the twentieth term is 56, so
    a + 19d = 56 .

    So now we have two equations in two unknowns, which we (you) can solve to find a and d. From there you should be able to complete the question.

    Hope this helps.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,672
    Thanks
    1489
    Quote Originally Posted by mastermin346 View Post
    The sum of the first term and seventh term of an arithmetic progression is 6.If the twentieth term is 56,find the term whose value first exceeds 400.
    t_n = a + (n - 1)d.


    t_1 = a + (1 - 1)d

    t_1 = a.


    t_7 = a + (7 - 1)d

    t_7 = a + 6d.


    t_1 + t_7 = 6

    a + a + 6d = 6

    2a + 6d = 6

    a + 3d = 3.


    t_{20} = 56

    a + (20 - 1)d = 56

    a + 19d = 56.


    Solving the equations simultaneously:

    a + 3d = 3 (1)

    a + 19d = 56 (2)


    (2) - (1):

    (a + 19d)-(a + 3d) = 56 - 3

    16d = 53

    d = \frac{53}{16}.


    So a + 3d = 3

    a + 3\left(\frac{53}{16}\right) = 3

    a + \frac{159}{16} = 3

    a = \frac{48}{16} - \frac{159}{16}

    a = -\frac{111}{16}.



    So to find the first value that is over 400...

    t_n = a + (n - 1)d

    400 = -\frac{111}{16} + \frac{53}{16}(n - 1)

    \frac{6400}{16} = -\frac{111}{16} + \frac{53}{16}(n - 1)

    \frac{6511}{16} = \frac{53}{16}(n - 1)

    \frac{6511}{53} = n - 1

    \frac{6564}{53} = n

    n = 123.85.


    So the first term that i going to be greater than 400 is t_{124}
    Last edited by Prove It; January 7th 2010 at 10:41 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jun 2009
    Posts
    220
    Thanks
    1
    EDIT: That step is most definitely not wrong
    Last edited by pomp; January 7th 2010 at 10:46 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,672
    Thanks
    1489
    Quote Originally Posted by pomp View Post
    That step is wrong (but I suppose that at least gives the OP some work to do )
    Doesn't \frac{6400}{16} = 400?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jun 2009
    Posts
    220
    Thanks
    1
    Quote Originally Posted by Prove It View Post
    Doesn't \frac{6400}{16} = 400?

    Woops! haha yes, sorry I went totally mathslexic for a second there.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Arithmetic progression
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: January 6th 2010, 02:10 AM
  2. Arithmetic Progression or Arithmetic Series Problem
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 8th 2009, 12:36 AM
  3. Sum of the sum of an arithmetic progression
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 26th 2009, 12:00 AM
  4. Replies: 8
    Last Post: March 23rd 2009, 07:26 AM
  5. Arithmetic Progression
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 1st 2008, 06:36 PM

Search Tags


/mathhelpforum @mathhelpforum