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Arithmetic progression

  1. May 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A finite arithmetic progression is given such that ##S_n>0## and ##d>0##. If the first member of the progression remains the same but ##d## increases by 2, then ##S_n## increases 3 times. If the first member of the progression remains the same but ##d## increases 4 times, then ##S_n## increases 5 times. Find ##d##.

    2. Relevant equations
    ##S_n=\frac{2a_1+d(n-1)}{2}\cdot n##

    3. The attempt at a solution
    ##\frac{2a_1+2d(n-1)}{2}\cdot n=\frac{2a_1+d(n-1)}{2}\cdot n \cdot 3##
    ##\frac{2a_1+4d(n-1)}{2}\cdot n=\frac{2a_1+d(n-1)}{2}\cdot n \cdot 5##

    When I try to solve this I get ##a_1=0## which is clearly not possible. Can somebody explain me what am I doing wrong?
     
    Last edited: May 4, 2016
  2. jcsd
  3. May 4, 2016 #2

    Samy_A

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    I also find ##a_1=0##.

    Are you sure about the way you interpret the exercise?

    Maybe it should be: "If ##d## is increased by 2, then ##S_n## is multiplied by 3", and similarly for the second one.
     
  4. May 4, 2016 #3
    Sorry for the confusion. I've updated my problem statement. The first part of the problem is: "If the first member of the progression remains the same but ##d## increases by 2, then ##S_n## increases 3 times." Everything else is the same.

    Now I got ##d=\frac{4}{3}##.
     
  5. May 4, 2016 #4

    Samy_A

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    So do I.
     
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