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.
Best Answer
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. $\frac{-1\pm\sqrt{14377}}{2}$