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Arithmetic Progression

  1. Mar 4, 2009 #1
    1. The problem statement, all variables and given/known data
    An arithmetic progression has n terms and a common difference of d. Prove that the difference between the sum of the last k terms and the sum of the first k terms is | (n-k)kd |.

    2. Relevant equations

    [tex]\begin{array}{l}
    {S_n} = \frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right] \\
    {u_n} = {a_1} + \left( {n - 1} \right)d \\
    \end{array}[/tex]


    3. The attempt at a solution
    I have no idea how to apply the "first 3 terms" and "last 3 terms" into the equation...
    Do I use [tex]{u_n}[/tex] as last term, and subsequently [tex]{u_{n - 1}}[/tex], [tex]{u_{n - 2}}[/tex] for last second and third term?
     
  2. jcsd
  3. Mar 4, 2009 #2

    tiny-tim

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    Hi fluppocinonys! :wink:
    Yes. :smile:
     
  4. Mar 5, 2009 #3
    I tried but still unable to solve it.
    Can you please hint me on how to start the question with?
    thanks
     
  5. Mar 5, 2009 #4

    tiny-tim

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    Hint: if n = 100, and k = 6, what is the difference between a1 and a95? :smile:
     
  6. Mar 5, 2009 #5
    a95 = a1 + 94d
    so,
    a95 - a1
    = a1 + 94d - a1
    = 94d
     
  7. Mar 5, 2009 #6

    tiny-tim

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    Yup! :biggrin:

    and then you … ? :wink:
     
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