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Math Help - Arithmetic series

  1. #1
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    Arithmetic series

    Hello all!

    Once again, I am stuck on a problem, and no matter how I twist it, I always end up in equations that have no solutions.

    Question:
    The sum of three consecutive members of an arithmetic progression is 2, and the sum of the squares of these members is \frac{14}{9}.

    My solution(done with the help of the book's hints)
    Let a be the "middle" number of the three numbers.
    a-d+a+d=4-(a-d)-(a+d)
    2a=4-2a
    4a=4
    a=1
    Then
    (a-d)^2+a^2+(a+d)^2=a^2-2ad+d^2+a^2+a^2+2ad+d^2
    3a^2+2d^2=\frac{14}{9}

    ans if we substitute a=1, we get

    2d^2=\frac{14}{9}-3
    2d^2=-\frac{7}{9}

    But then the equation has no solutions.

    All help is appreciated. Anyway, why is it so that some equations do not end up having solutions?
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  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
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    Quote Originally Posted by Coach View Post
    Hello all!

    Once again, I am stuck on a problem, and no matter how I twist it, I always end up in equations that have no solutions.

    Question:
    The sum of three consecutive members of an arithmetic progression is 2, and the sum of the squares of these members is \frac{14}{9}.

    My solution(done with the help of the book's hints)
    Let a be the "middle" number of the three numbers.
    a-d+a+d=4-(a-d)-(a+d)
    2a=4-2a
    4a=4
    a=1
    where did you get the 4 from?

    If you have 3 successive terms with difference d it should go...

    (a-d) + a + (a+d) = 2

    => a + a + a - d + d = 2

    => 3a = 2

    => a = \frac{2}{3}

    And then,

    (a-d)^2 + a^2 + (a+d)^2 = \frac{14}{9}

    => \left ( \frac{2}{3} - d \right )^2 + \frac{4}{9} + \left ( \frac{2}{3} + d \right )^2 = \frac{14}{9}

    => \frac{4}{3} + 2d^2 = \frac{14}{9}

    => d^2 = \frac{1}{9}

    => d = \frac{1}{3}

    So your sequence is \frac{1}{3}, \frac{2}{3}, \frac{3}{3}.

    Add them up, it comes to 2.

    Add the squares up...

    \frac{1}{9}+\frac{4}{9}+\frac{9}{9} = \frac{14}{9}
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  3. #3
    Member
    Joined
    Jul 2007
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    Thank you so much!

    I thought that the average of the first and the second term equals the "middle term". And hence \frac{(a-d)+(a+d)}{2}=2 and (a-d)+(a+d)=4,
    but now I realized where I made a mistake.
    Thanks a lot!
    Quote Originally Posted by Deadstar View Post
    where did you get the 4 from?

    If you have 3 successive terms with difference d it should go...

    (a-d) + a + (a+d) = 2

    => a + a + a - d + d = 2

    => 3a = 2

    => a = \frac{2}{3}

    And then,

    (a-d)^2 + a^2 + (a+d)^2 = \frac{14}{9}

    => \left ( \frac{2}{3} - d \right )^2 + \frac{4}{9} + \left ( \frac{2}{3} + d \right )^2 = \frac{14}{9}

    => \frac{4}{3} + 2d^2 = \frac{14}{9}

    => d^2 = \frac{1}{9}

    => d = \frac{1}{3}

    So your sequence is \frac{1}{3}, \frac{2}{3}, \frac{3}{3}.

    Add them up, it comes to 2.

    Add the squares up...

    \frac{1}{9}+\frac{4}{9}+\frac{9}{9} = \frac{14}{9}
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