Results 1 to 10 of 10
Like Tree1Thanks
  • 1 Post By emakarov

Math Help - Geometric progression and arithmetic progression problem.

  1. #1
    Newbie
    Joined
    Aug 2012
    From
    Albania
    Posts
    9

    Talking Geometric progression and arithmetic progression problem.

    Hi,
    I'm having difficulties with a problem with arithmetic and geometric progression i appreciate your help.
    Here's the problem:
    a,b,c form a geometric progression and a+b,b+c,a+c form an arithmetic progression.
    I need to find the q=b/a=c/b.
    Thank you for your help!!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,527
    Thanks
    773

    Re: Geometric progression and arithmetic progression problem.

    Express b and c through a and q, write the equation saying that the difference between the third and first terms of arithmetic progression is twice the difference between the second and first terms, simplify and solve for q. Don't forget special cases a = 0 and q = 1.
    Thanks from DarkAngle
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2012
    From
    Albania
    Posts
    9

    Re: Geometric progression and arithmetic progression problem.

    b=a*q
    c=a*q^2
    (a+c-a+b)=2(b+c-a+b)
    c+b=4b+c-a
    3b=a
    3aq=a 3q=a/a 3q=1 q=1/3 Is that correct?
    Thank you for youe help!!!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,527
    Thanks
    773

    Re: Geometric progression and arithmetic progression problem.

    Quote Originally Posted by DarkAngle View Post
    (a+c-a+b)=2(b+c-a+b)
    This should be a + c - a - b = 2(b + c - a - b).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Aug 2012
    From
    Albania
    Posts
    9

    Re: Geometric progression and arithmetic progression problem.

    Yes i forgot thanks for the correction
    a+c-a-b=2(b+c-a-b)
    c-b=2c-2a
    -c-b=-2a ---> c+b=2a
    aq^2+aq=2a
    a(q^2+q)=2a ---> q^2+q-2=0 1-4*-2=9 ---> (-1+3)/2=1 (-1-3)/2=-2
    seems this equation has two solution
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,527
    Thanks
    773

    Re: Geometric progression and arithmetic progression problem.

    Quote Originally Posted by DarkAngle View Post
    (-1+3)/2=1 (-1-3)/2=-2
    seems this equation has two solution
    That's right. Check that a+b, b+c, a+c indeed form an arithmetic progression for each of those q. Also note that you divided both sides of a(q^2+q)=2a by a, which is possible only if a ≠ 0. What happens when a = 0?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Aug 2012
    From
    Albania
    Posts
    9

    Re: Geometric progression and arithmetic progression problem.

    If a=0 then we don't have an equation at all.a,b,c is an geometric progression which mean a=0 leads to b=0 and c=0.
    b=a*q a=0
    b=0*q=0
    c=a*q^2
    c=0*q^2=0
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Aug 2012
    From
    Albania
    Posts
    9

    Re: Geometric progression and arithmetic progression problem.

    The first term must be different to 0 to have an geometric progression the fact that it is a geometric progression means a different to 0
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,527
    Thanks
    773

    Re: Geometric progression and arithmetic progression problem.

    Quote Originally Posted by DarkAngle View Post
    The first term must be different to 0 to have an geometric progression the fact that it is a geometric progression means a different to 0
    This depends on the definition, though Wikipedia, MathWorld and the Encyclopedia of Mathematics do not seem to require that a ≠ 0, only that q ≠ 0. If zero elements are not allowed, then dividing by a is justified and there are two solutions for q. If zero elements are allowed, then q can be any nonzero number.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Aug 2012
    From
    Albania
    Posts
    9

    Re: Geometric progression and arithmetic progression problem.

    In my class and my book it says that every term must be different from 0
    if a=0 then second term b=0
    q=b/a ---> q=0/0 you can't divine by zero and that leads to nonexistence of q
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: June 24th 2010, 08:20 AM
  2. Arithmetic progression problem
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: January 8th 2010, 10:23 PM
  3. Arithmetic & Geometric Progression
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 20th 2009, 08:00 PM
  4. Replies: 8
    Last Post: March 23rd 2009, 07:26 AM
  5. Arithmetic & Geometric Progression
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 5th 2008, 07:16 AM

Search Tags


/mathhelpforum @mathhelpforum