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Arithmetic Series problem

  1. Dec 31, 2013 #1
    1. The problem statement, all variables and given/known data
    An arithmetic series consists of 2n terms. Which are the two middle terms of the series? If the first term is a and the last term is b, find the middle terms and the sum of the series.

    2. Relevant equations

    3. The attempt at a solution

    I'm having problems finding out which terms are the middle terms. For a previous question all I did was find the average of the 1st and last term, but how would I know which term it is, I would know the value though. Please give me some guidance.
  2. jcsd
  3. Dec 31, 2013 #2
    With an odd number of terms, the middle term is simply (a+b)/2.
    If you draw a series with an even number of terms on a number line, the point (a+b)/2 will be in the middle of the two terms you want.
    It ill be easy to find those points, if you compute the distance between two terms in the series first.
  4. Dec 31, 2013 #3
    Is there any way to do it algebraically?
  5. Dec 31, 2013 #4


    Staff: Mentor

    You can write the series like so:
    a1 + a2 + ... + an + an+1 + an+2 + ... + a2n

    From the given information, a1 = a and a2n = b. Clearly, the middle terms are an and an+1. You'll need to incorporate the information that this is an arithmetic series. I'm hopeful that this is enough to get you started.
  6. Dec 31, 2013 #5
    Hmm I think I see it, I had a problem counting like from 1 to 2n... So because 2n is even you just put an even number of terms. I was thinking how do you know 2n is more than n and n+1, what if n was 2, n+1 and n+2 would be 3 and 4 and 2n is also 4... or that doesn't matter?
  7. Jan 1, 2014 #6


    Staff: Mentor

    It's reasonable to assume that n ≥ 1, so 2n will always be larger than n. If n = 1, then n + 1 = 2n, but if n > 1, then 2n will be larger than n + 1.

    If your series consists of only two terms (i.e., n = 1), then it's not very interesting. If you feel you need to, you can handle that as a special case.
  8. Jan 1, 2014 #7
    Oh thanks!
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