In an arithmetic series formula, can the n be negative? I.e., if you're looking for how many terms you need to sum in 2 + 5 + 8 + ... to get to say (for example) greater than 243, what if the quadratic you end up solving gives you a negative number? How would that work and why is it an acceptable answer? You can't have a negative number of terms, so why is it considered an appropriate answer?


  • $\begingroup$ I don't think it's an appropriate answer in this context. $\endgroup$ – Akiva Weinberger Oct 17 '15 at 23:14
  • $\begingroup$ When would it be an appropriate answer? $\endgroup$ – user164403 Oct 18 '15 at 0:59

If you think that when you are solving the quadratic equation you are doing so out of context of the question. You can always get solutions to a quadratic (whether real/complex, positive/negative) but when you get them you must check whether they make sense in the context of what you're doing. Really in your example we don't want to just solve $\frac{n}{2}(2\times2+(n-1)3)=243$ but the system: $$ n\in\mathbb{Z}\;\land n\geq0\;\; \land\; \frac{n}{2}(2\times2+(n-1)3)=243 $$ (Where $\land$ is the logic symbol for 'and', so they must all be true). This seems reasonable since, as you say, we haven't defined what we mean by a sequence with negative terms; so $n>0$, clearly it must be an integer and the quadratic comes from the sum. Now if we solve the quadratic and that gives values of $n$ that aren't positive integers then we don't solve the system and thus they are not solutions. If you get two integers solutions, one positive and the other negative, then we only consider the one that satisfies all conditions, the positive one.

Further to comment

You can take the absolute value of the negative solution for $n$ and put it into: $$ \frac{n}{2}(2(d-a)+(n-1)d) $$ and that'll give the same sum as the positive value of $n$. So in your example you'd put in $a=12$ and $d=4$ with $n=13$ and the sum would be 204. Which can then be interpreted as a series that starts from $d-a$ and then has common difference $d$.

The $40$ comes from the fact when you work out the values of this new arithmetic progression you get the sequence (up to $n=13$): $$(-8, -4, 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40)$$ which sums to 204. They have just reversed the series by writing it as a descending sum rather than an ascending one, though clearly they sum to the same thing.

  • $\begingroup$ One of my questions in my worksheet asks: "How many terms of 12 + 16 + 20 + ... will give a sum of 208? This problem leads to a quadratic equation with two integer roots. Give an interpretation of both roots." I solved and got n = 8, and n = -13. Clearly in this context n = -13 does not work, yet the worksheet gives an interpretation of the negative answer as follows: (if n = -13, 40 + 36 + 32 + 28 + 24 + ... + 0 - 4 - 8 = 208). Could you explain this? $\endgroup$ – user164403 Oct 18 '15 at 1:24
  • $\begingroup$ @user164403 I'm not entirely sure where they got that from, apart from just finding another series that happened to work. I was looking into a bit before answering and so I'll add a bit to my answer. $\endgroup$ – Jay Oct 18 '15 at 1:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.