Middle Terms in an Arithmetic Series

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In an arithmetic series with 2n terms, the two middle terms are identified as an and an+1, where a is the first term and b is the last term. To find these middle terms, one can calculate the average of the first and last terms, (a+b)/2, which lies between them. The discussion emphasizes that since 2n is even, there will always be a clear distinction between the middle terms, and n must be greater than or equal to 1 to ensure 2n exceeds n+1. A special case is acknowledged when n equals 1, resulting in only two terms, which is less complex. Understanding the structure of the series is crucial for determining the middle terms and their values.
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Homework Statement


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.

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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.
 
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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.
 
Is there any way to do it algebraically?
 
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.
 
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?
 
lionely said:
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?

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.
 
Oh thanks!
 
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