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Given the sum of a series and a term how would you find Tn?

  1. Jan 26, 2015 #1
    Hi all, given the sum of a series and a single term how would one find the nth term? Any help appreciated.
     
  2. jcsd
  3. Jan 26, 2015 #2

    HallsofIvy

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    You can't. That's obviously impossible. There exist an infinite number of series having a given specific term and converging to a give value.

    Suppose you are given the term "[itex]a_i[/itex]", for i a specific number, and that the sum is "S". Choose "A" to be any number and write the geometric sequence that converges to [itex]S- a_i[/itex]. Then the series consisting first term A, the rest of the geometric series with "[itex]a_i[/itex]" stuck in a the ith place, has sum S.
     
  4. Jan 26, 2015 #3
    It's arithmetic, never mind I figured it out, apologies for wasted time.

    t6=32
    s9=234
    32=a+5d
    d=(32-a)/5

    234=0.5*9(2a+8((32-a)/5))
    a=2
    d=6
    Tn=6n-4
     
  5. Jan 27, 2015 #4

    HallsofIvy

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    In your original post you said "given the sum of a series and a single term". What are "t6", "s9", "a" and "d".
     
  6. Jan 27, 2015 #5
    a is the first term, d is the difference between terms, t6 is 6th term, s9 is sum of the first 9 terms
     
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