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So, it's been a long time since I've studied math, so I'm having more trouble with this problem than I thought I would as for some help. I have an arithmetic sequence $0,...,99$ with the difference being $1$. Basically I have numbers $0$ to $99$. The sum of this sequence is $4950$. If I then say that the sum $1797 = 0,...,n$, how would I find $n$?

I've gotten to the point in my equation where $2(1797) = n * (n+1)$ but I don't know where to go from here.

Also, obviously, there is no whole number solution to this particular issue, however there is a rational one and that's what I am looking for.

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It's a quadratic equation. And so it will have 2 roots. Solve it using the quadratic formula. If the result is not rational, then too bad. – Gautam Shenoy Nov 15 '12 at 16:48
Perhaps "rational" is not the right word... It has been awhile. – davethegr8 Nov 16 '12 at 15:41
up vote 1 down vote accepted

Use simple formula for quadratic equations. Re-writing your equation you get $n^2 + n - 2*1797 = 0$. The number by $n^2$ is customarily named $a$, the one by $n$ is $b$ and the third one $c$.

There are two solutions given by: $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ which gives us both solutions i.e. $\frac{-1\pm\sqrt{14377}}{2}$

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It doesn't make sense to say $0,...,\frac{-1\pm\sqrt{14377}}2$, though. What does that even mean? – Cameron Buie Nov 15 '12 at 17:03
@Cameron this would be a useful tool, though, if the problem were slightly modified to ask, "What is the largest $n$ for which $1+2+3+\cdots + n \le 1797$?" – Rick Decker Nov 15 '12 at 17:18
True, @Rick, or if we wanted the least $n$ such that $1+2+3+\cdots+n\geq 1797$. – Cameron Buie Nov 15 '12 at 17:23
@CameronBuie That was the colsest I could get to: "obviously, there is no whole number solution to this particular issue, however there is a rational one and that's what I am looking for." I'm closer then any $\varepsilon$ you pick to some rational number ;) – Golob Nov 15 '12 at 23:22
Ok, so rational's not the right term. But both of those examples are the kind of thing I needed. – davethegr8 Nov 16 '12 at 15:42

You are right you want to solve $2\cdot 1797=n(n+1)$, which is $n^2+n-3594$. You can look to factor this or use the quadratic equation to get $$n=\frac{-1\pm\sqrt{1+4\cdot 3594}}2$$ Which has solutions about $-60.5$ and $59.5$. Neither of these is rational as $14377$ is not a square.

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It's been awhile... basically I was looking for the larger one and I guess rational isn't the right term to use. – davethegr8 Nov 16 '12 at 15:42

We typically use $n$ to denote (non-negative) integers, but that isn't a huge deal.

If you're talking about an arithmetic sequence with difference $1$ starting at $0$, then every number in the sequence will be a non-negative integer. There's no avoiding that.

If we're not constrained to a difference of $1$, then you may as well just take the sequence to be $0,1797$. There's really no need to to get into sequences with non-integer rational difference.

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Sorry, I'm having a hard time figuring out how to describe this problem. Basically, if $4950=sum(1...99)$ then $1797=sum(1...x)$ – davethegr8 Nov 15 '12 at 16:54
I understand what you're trying to do. Here's the kicker, though: what does "..." mean in the context of $1,...,x$ for arbitrary $x$? There's a natural interpretation when $x$ is a positive integer--namely, that we're talking about an arithmetic sequence with difference $1$. You've already noticed, though, that there's no integer solution to $x^2+x=2(1797)$--in fact, there isn't even a rational solution (not that it would make sense, anyway). – Cameron Buie Nov 15 '12 at 16:57

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