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Which is the 9th term where do i stop adding?

  1. Mar 7, 2005 #1
    find the sum of the first 9 terms in the series

    9+11+13+15+17+19+............

    is the 9 t0 or t1?
    what is the 9th term?
    ...........21+23+25+27

    25 or 27??
     
  2. jcsd
  3. Mar 7, 2005 #2

    chroot

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    The first nine terms are... the first nine terms.

    9
    11
    13
    15
    17
    19
    21
    23
    25

    Count them.

    - Warren
     
  4. Mar 7, 2005 #3

    dextercioby

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    And you have two ways to add them:brute force (simply add the 9 #-s) or elegantly,using the formulas for an arithmetic progression.

    Daniel.
     
  5. Mar 7, 2005 #4
    ok I got that part thanks, the next part says the sum of the first 5 terms can be found by S2=[n(t1+t5)]/2

    create a formula to find the sum of the first 8 terms I just replace t5 in the above equation with t8 is this correct?

    The next part says create a formula to find the sum of the first n terms how do u do this?
     
  6. Mar 7, 2005 #5

    dextercioby

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    Is that "n" the step...?I think so.There's a nice and very old proof for the general formula.Do you know it (has it been taught to you at school...?)?

    Daniel.
     
  7. Mar 7, 2005 #6

    dextercioby

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    Nope,i checked.That "n" is not the step,but the # of terms.Sorry.

    Daniel.

    P.S.So in your example that "n" should be first 5,then 8,then arbitrary "n".
     
  8. Mar 7, 2005 #7
    The common formula for calculating arithmetic series should be in your book, or you should have discussed it? Have you ever seen a general formula?
     
  9. Mar 7, 2005 #8
    yes i know n is the number of terms

    A formula to find the sum of the first 8 terms would be

    S2=[n(t1+t8)]/2 where n=8 and t1 and t8 are taken from the series in my first post .

    How do u create a formula to find the sum of the first n terms?
    Yes ive seen the one in my book there are two sn= (n/2)(a+tn) when d is unknown
    and sn=(n/2)[2a+(n-1)d] when tn is unknown? Which one do I put for this question because I think from the previous part that shows how to find the sum of 5 terms and my formula for finding the sum of 8 terms needs to be used to derive sum of first n terms....
     
    Last edited: Mar 7, 2005
  10. Mar 7, 2005 #9

    dextercioby

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    There's the simple proof.U can define an arithmetical progression through the recurrence relation:
    [tex] a_{1}+a_{n}=a_{2}+a_{n-1} [/tex]

    Use that and the trick
    [tex] S=a_{1}+a_{2}+...+a_{n-1}+a_{n} [/tex]
    [tex] S=a_{n}+a_{n-1}+...+a_{2}+a_{1} [/tex] And simply add:
    -----------------------------------------------
    [tex]2S=(a_{1}+a_{n})+(a_{2}+a_{n-1})+...+(a_{n-1}+a_{2})+(a_{n}+a_{1}) [/tex]

    Use the recurrence relation and a simple counting to write
    [tex] 2S=n(a_{1}+a_{n}) [/tex]
    [tex]S=\frac{n}{2}(a_{1}+a_{n}) [/tex]

    Simple,huh...?

    Daniel.
     
  11. Mar 7, 2005 #10
    Yes thanks I got it! :smile:
     
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