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Math Help - Sequence and Series (guide me, thanks!)

  1. #1
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    Sequence and Series (guide me, thanks!)

    In an arithmetic progression, the sum of the first n even number terms exceeds the sum of the first n odd number terms by 3n. Find the common difference of this series.

    my working:

    Sigma a(2r) - Sigma a(2r-1) = 2 . upper limit of sigma - n , lower limit - 1
    Note that 2r and 2r-1 are both subscripts.

    I have no idea what to do next, help me please, thank you!
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  2. #2
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    Joined
    Sep 2008
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    I seem to have done it but not with the working shown in the first post.

    Sum of first n even number terms = (a+d) + (a+3d) + (a+5d) + ... + a + ( 2n-1 ) d -----------1
    Sum of first n odd number terms = (a) + ( a+2d) + (a+4d) +... +a + ( 2n-2)d ---------------- 2

    Notice that the series has n terms, therefore , the number a of in both sum are equal.
    However, the number of d in both sum are different such that 1 is more than 2 for nd ( since a2-a1 = d a4-a3 =d a6-a5=d .... an - an-1 = d and there is a total of n terms, therefore , the sum of first n even number terms exceeds the sum of first n odd number terms by nd. )

    Given that 1 -2 = 3n and what I found , 1 - 2 = nd ,
    Therefore, 3n = nd
    d = 3

    Any better way or any correction in my working here?

    Thanks.
    Last edited by ose90; February 26th 2009 at 03:17 AM.
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