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Arithmetic series

  1. Jun 6, 2006 #1

    Aya

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    OK, these questions are supposed to be really easy, but I keep geting the wrong answer and I don't understand why, could someone pleas look at my work and tell me what I am doing wrong?

    Find the indicated sum of each arithmetic series

    S15 for 4 + 11 + 18 + ...

    Sn= n/2 [ 2a + ( n - 1 ) d ]
    Sn= 15/2 [ 2(4) + (15 - 1 ) 7 ]
    Sn= 7.5 [ 8 + 98 ]
    Sn= 795

    Find t8 for the following sequence

    10 000 , -5 000 , 2 500 , -1 250 ,

    So...

    tn= ar ^ n-1
    tn= 10 000 (-0.5) ^ n-1

    t8= 10 000 (-0.5) ^ 7
    = 78.125


    Sometimes the answers in the back of the book are wrong, so am I wrong or are they??
     
  2. jcsd
  3. Jun 6, 2006 #2

    HallsofIvy

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    Staff Emeritus
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    Presumably you not that if an has initial value a0 and "constant difference" x, then an= a0+n(xn. Furthermore, the sum of the first n terms of an arithmetic series is [itex]a_0+ a_n= (a_0+ x_n)n/2[/itex]. That is, the sum is n times the "average" of the first and last terms of the sequence.
    In the first problem, a0= 4 and r= 11- 4= 7. The first term is 4 and the 15th term is 4+ 14(7)= 4+ 98= 102. The sum is (4+ 102)/2= (106)/2= 53.
     
  4. Jun 6, 2006 #3

    VietDao29

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    Homework Helper


    Uhm , the sum should be 53 times 15 (terms of the sequence) = 795. You seem to have dropped out the 15 somewhere. :smile:
    -------------
    @Aya: You did the first problem correctly :wink:.
    The second one, however is wrong. By the way, are you trying to find the 8-th term of the series? Or the sum of the first 8 terms?
    Assuming that you want to find the 8-th term, there's a small error in your work: you seem to forget a minus sign. :smile:
     
  5. Jun 7, 2006 #4

    Aya

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    ^ ok, thanks for all your help
     
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