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Series and Sequence,

  1. Oct 21, 2005 #1


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    Lol it's quite easy but these questions r annoying me.
    The 20th term of an arithmetic series if 131 and sum of the 6th to 10th term inclusive is 235, find sum is the first 20 terms
    131= a + 19 d 5(2a+ 9d) -3(2a+5d)=235
    well i just went on and solve simutalneosly and got the wrong answer for S(20), i don't know how to make it at hte butotm.
    2) the sum of 50 terms of an arithmetic is 249 and sum of 49 terms is 233, find 50th term of the series.
    249= 50a + 1225d and 233= 49/2 (2a+48d)
    and well got the wrong answer aagain.
    3) prove [itex]T_n = S_n - S_n_-_1[/itex]
    I have no idea how to do that. Thanks
  2. jcsd
  3. Oct 22, 2005 #2


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    Homework Helper

    I have bolded the wrong equation in your post. In fact the sum of the sixth term to the tenth term is:
    S6 - 10 = S10 - S5 = [tex]5(2a + 9d) - \frac{5}{2}(2a + 4d) = 235[/tex]
    S6 - 10 = S10 - S6

    You are complicating the problem...
    S50 = a1 + a2 + a3 + ... + a49 + a50
    S49 = a1 + a2 + a3 + ... + a49
    So what's a50?

    I have no idea what Tn is... Is that the n-th term of the series?
    Viet Dao,
  4. Oct 22, 2005 #3
    your second equation is wrong. it should read 5a + 35d = 235; after that you can solve the system formed withthis equation et your first equation to find a and d
    After that it should be easy to find the sum of the first 20 terms.
  5. Oct 23, 2005 #4
    Concept of mean is often helpful in questions on arithmetic sequences.
    Mean of n first terms of such sequence is
    amean = Sn/n
    If n is odd, amean is one of the terms of the sequence.
    Which one?
    Knowing that
    a6 + a7 + a8 + a9 + a10 = 235
    you can determine the value of a?.
    Now you can find a1 from a? and a20 in one step.
    a1 and a20 lead you straight to S20.
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