Results 1 to 6 of 6

Math Help - arithmetic progression

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    4

    arithmetic progression

    how do i find common difference if the first term is 3 and the seventh term is twice the third term. bit of help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    11
    Awards
    1
    Quote Originally Posted by finnkeers View Post
    how do i find common difference if the first term is 3 and the seventh term is twice the third term. bit of help.
    Hi finnkeers,

    Use a_n=a_1+(n-1)d

    Let x=a_3


    a_3=a_1+(n-1)d

    x=3+(3-1)d

    x=3+2d

    {\color{red}d=\frac{x-3}{2}}

    Do a similar calculation using:

    Let a_7=2x


    a_7=3+(7-1)d

    2x=3+6d

    {\color{red}d=\frac{2x-3}{6}}

    Now set the two d's equal to each other and solve.
    Last edited by masters; May 19th 2010 at 05:34 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2010
    Posts
    4
    sorry but now i have two variables so how do i solve for d by getting rid of x
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A riddle wrapped in an enigma
    masters's Avatar
    Joined
    Jan 2008
    From
    Big Stone Gap, Virginia
    Posts
    2,551
    Thanks
    11
    Awards
    1
    Quote Originally Posted by finnkeers View Post
    sorry but now i have two variables so how do i solve for d by getting rid of x
    Well, let's see.

    d=\frac{x-3}{2} and d=\frac{2x-3}{6}

    Setting them equal to each other, we have

    \frac{x-3}{2}=\frac{2x-3}{6}

    x=6

    If x = 6, then substituting into either equation with d, we get

    d = \frac{6-3}{2} =\boxed{\frac{3}{2}}

    d = \frac{2(6)-2}{6}=\frac{9}{6}=\boxed{\frac{3}{2}}

    And there you have it!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member BobBali's Avatar
    Joined
    May 2010
    From
    Tanzania
    Posts
    61
    first term (u1) = 3

    u1 + 6d = 7th term But 7th term = 2(u1+2d)

    set equations equal to one another:

    3 + 6d = 2(3+2d)
    3+6d = 6+4d
    6d-4d = 6-3
    2d= 3

    Therefore, d=3/2
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Mar 2010
    Posts
    4
    thanks everyone. much appreciated.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Arithmetic Progression or Arithmetic Series Problem
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 8th 2009, 12:36 AM
  2. Arithmetic progression help..
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: April 6th 2009, 11:39 AM
  3. Replies: 8
    Last Post: March 23rd 2009, 07:26 AM
  4. arithmetic progression--need help 2
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 14th 2009, 10:45 PM
  5. Arithmetic Progression X.x
    Posted in the Algebra Forum
    Replies: 9
    Last Post: May 5th 2008, 06:12 AM

Search Tags


/mathhelpforum @mathhelpforum