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I am trying to understand the derivation of the formula for the sum of an arithmetic sequence of the first $n$ terms.

I do not understand what rules or reasoning allow two sequences to be added in reverse order to eliminate the common difference $d$ and arrive at the conclusion that the sum of an arithmetic sequence of the first $n$ terms is one half $n$ times the sum of the first and last terms. This seems to be a contrived way to eliminate the common difference from the expanded based on some unexplained knowledge of $d$ and arithmetic sequences in general.

I have researched this question in maths textbooks and online and each time the derivation is presented I cannot seem to find an explanation as to why it would be evident to a mathematician that by adding the sequences they would derive the formula.

The background.

The derivation of the formula as explained in many textbooks and online sites is as follows.

  1. To find the sum of an arithmetic sequence for the first $n$ terms $S_n$, we can write out the sum in relation to the first term $a_1$ and the common difference $d$.

$$ S_n = a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + ... + a_n $$

  1. It is also possible to write the sequence in reverse order in relation to the last term $a_n$.

$$ S_n = a_n + (a_n - d) + (a_n - 2d) + (a_n - 3d) + ... + a_1 $$

  1. When we add these sequences together we derive the formula for the sum of the first n terms of an arithmetic sequence.

$$ \begin{array}{r} S_n = a_1 + (a_1 + d) + (a_1 + 2d) + (a_1 + 3d) + \ldots + a_n \\ + \,S_n = a_n + (a_n - d) + (a_n - 2d) + (a_n - 3d) + \ldots + a_1 \\ \hline 2S_n = (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) + (a_1 + a_n) \ldots \end{array} $$

  1. Because there are $n$ many additions of $(a_1 + a_n)$ the lengthy sum is simplified as $n(a_1 + a_n)$ and solving for $S_n$ we arrive at the formula.

$$ S_n = \frac{n}{2}(a_1 + a_n) $$

Unfortunately I can't seem to find the reasoning in any of these explanations as to why the two sequences (ordinary order and reverse) were added. It makes sense to me that they were added but not why this was the next logical step when deriving the formula.

The question.

Why were the two sequences added to derive the formula and what does that show about the nature of arithmetic sequences?

In my attempt to figure this out I noted that by studying many sequences we can see that the ratio of the sum of the sequence for the first $n$ terms $S_n$ and the sum of the first and last terms $(a_1 + a_n)$ is always $\frac{n}{2}$ for any arithmetic sequence. So possibly it could be said by induction that if for any arithmetic sequence it is true that:

$$ \frac{S_n}{a_1 + a_n} = \frac{n}{2} $$

Then it must also be true that:

$$ S_n = \frac{n}{2}(a_1 + a_n) $$

However, to me this still doesn't explain why the derivation decides to add the two sequences.

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  • $\begingroup$ "and then point 3 decides to conclude on the basis of some unclear reason that by adding the two sequences we can eliminate $d$..." What is unclear about it? Do you not see that $(a_1+d)+(a_n-d) = (a_1+a_n)+(d-d)=a_1+a_n$? $\endgroup$
    – JMoravitz
    Aug 16, 2020 at 2:11
  • $\begingroup$ @JMoravitz yes this is very clear. My question (which I will update to be clearer) is why someone deriving the formula (for the first time for example) knew to write the sequences in both ways and then add them. Is it simply the case that it's obvious that doing this eliminates the common difference (d) and eliminating variables is desirable or is there a well known fact or property that hints that common difference will be eliminated? $\endgroup$
    – b_n
    Aug 16, 2020 at 5:06
  • $\begingroup$ "I can't seem to find the reasoning in any of these explanations as to why the two sequences (ordinary order and reverse) were added." Because it works. There doesn't need to be any more reason than that. " It makes sense to me that they were added but not why this was the next logical step when deriving the formula" Because it works. That's all that matters. You don't need to know how the figured it out. You just need to know what they did. And that's completely obvious once it's been pointed out, isn't. $\endgroup$
    – fleablood
    Aug 16, 2020 at 6:03
  • $\begingroup$ Well, my thinking was when I was told to add all the numbers from $1$ to $100$ that I quickly saw that numbers that ended with $k$ and $10-k$ were easier to add up and that the came in equal pairs and I could add those and mulitply. I first tried to do in small groups (add $41$ to $49$, $42+48$. But then it was easy to see I could just make one long chain and make it a single multiplication. $\endgroup$
    – fleablood
    Aug 16, 2020 at 6:10
  • $\begingroup$ " Is it simply the case that it's obvious that doing this eliminates the common difference (d) and eliminating variables is desirable or is there a well known fact or property that hints that common difference will be eliminated?" Well, ... yeah... I guess so. If you ever had the task of having to add up a list of twenty numbers without a calculator, this becomes pretty obvious pretty quick. It's hard for me to imagine someone not coming up with shortcuts. Although to come up with a single short cut and prove it always works takes a certain amount of acuity. $\endgroup$
    – fleablood
    Aug 16, 2020 at 6:13

2 Answers 2

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Commutativity of addition lets you permute the sum of two addends. By induction on the number of addends you can extend this to any finite number of addends. Associativity then lets you group them however you want.

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  • $\begingroup$ I believe you're addressing why adding the two sequences eliminates the common difference (d). I would like to know why someone looking to derive this formula would take the method of writing the sum in two ways and then adding them. Do they already know that the d will be eliminated or is this discovered by adding the sequences? If they already know the d will be eliminated, what fact or property told them the d was irrelevant in the sequence? $\endgroup$
    – b_n
    Aug 16, 2020 at 5:02
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    $\begingroup$ How you find it is very different from how you prove it. Finding it is often a result of playing around. It would be natural to take a sequence, reverse it, and add term by term. That would give a constant, which leads to the usual formula. Having found that, you try to find a way to prove it is true. Proofs in a text do not describe how the formula was found, they just prove it is true. Many people play with math, often with sequences, and some stumble across an interesting result. We see the results of this playing here sometimes. $\endgroup$
    – Ross Millikan
    Aug 16, 2020 at 5:10
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    $\begingroup$ I don't find this one mysterious at all. You have a sequence increasing by $d$, so if you add it to a sequence decreasing by $d$ you get a constant. Constants are always interesting. $\endgroup$
    – Ross Millikan
    Aug 16, 2020 at 5:11
  • $\begingroup$ Thanks @Ross Millikan you answered my question by explaining that i. it is natural reasoning to add a sequence increasing by some value n to a sequence decreasing by n because this will result in a constant which can be used in the remainder of the derivation, ii. the steps for derivation are depart from subsequently proving the derivation is correct and derivations in general would use natural steps like adding the two sequences. Please add this as an answer or I am happy to write this as an answer and quote your comments. $\endgroup$
    – b_n
    Aug 16, 2020 at 6:00
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Think back to the sum of triangle numbers. To add the first n numbers, you take the first and last and combine them to get n+1, then you do the same for the second and second to last to get n+1. For a visual of this process, look at https://en.wikipedia.org/wiki/Triangular_number. Here in your question, your triangle is a little lopsided and doesn't come to a point, but the idea remains the same :).

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