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Arithmetic series

  1. Jun 4, 2012 #1
    1. Write out in full and determine the sum of the finite arithmetic series

    r = 9
    Ʃ 5r 5,10,15,20,25,30,35,40 <-- until 9
    r=1

    But how do I determine the sum of the finite arithmetic series? I forget the formula :/
     
  2. jcsd
  3. Jun 4, 2012 #2

    Mentallic

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    Homework Helper

    It's not too hard to find out the formula for yourself!

    Start with some general arithmetic series starting with the value 'a' and each successive term has a difference 'd'.

    So our summation would go like this:

    [tex]S=a + (a+d) + (a+2d) +(a+3d)+ ... + (a+(n-1)d)[/tex] which is a total of n terms.

    Now let's look at the same summation, but this time we begin with the last term first, the second last goes second, etc.

    [tex]S=(a+(n-1)d)+(a+(n-2)d)+...+(a+d)+a[/tex]

    And now let's add both these summations together. The first adjacent terms will be [itex]a+(a+(n-1)d)=2a+(n-1)d[/itex], the second will be [itex](a+d)+(a+(n-2)d)=2a+(n-1)d[/itex] and consequently every sum of each two adjacent terms will add up to this, so what we end up with is:

    [tex]2S=(2a+(n-1)d)+(2a+(n-1)d)+...+(2a+(n-1)d)[/tex] which has a total of n terms, so we can simplify it into

    [tex]2S=n(2a+(n-1)d)[/tex]

    [tex]S=\frac{n}{2}\left(2a+(n-1)d\right)[/tex]

    Which is the formula you're looking for.
     
  4. Jun 4, 2012 #3

    HallsofIvy

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    A crucial point about arithmetic series is that the average of all numbers in an arthmetic series is the same as the average of the first and last terms.

    Here the first term is 5(1)= 5 and the last term is 5(9)= 45 (I don't know why you stopped the series in your post at 40) so the average term is (5+ 45)/2= 50/2= 25.
     
    Last edited: Jun 5, 2012
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