Problem (Algebra):
Let $T=72 \sqrt{2}$, and let $K=\left(\frac{T}{12}\right)^{2}$. In the sequence $0.5,1,-1.5,2,2.5,-3, \ldots$, every third term is negative, and the absolute values of the terms form an arithmetic sequence. Compute the sum of the first $K$ terms of this sequence.

Solution:
The general sequence looks like $x, x+d,-(x+2 d), x+3 d, x+4 d,-(x+5 d), \ldots$ The sum of the first three terms is $x-d$; the sum of the second three terms is $x+2 d$; the sum of the third three terms is $x+5 d$, and so on. Thus the sequence of sums of terms $3 k-2,3 k-1$, and $3 k$ is an arithmetic sequence. Notice that $x=d=0.5$ and so $x-d=0$. If there are $n$ triads of terms of the original sequence, then their common difference is 1.5 and their sum is $n \cdot\left(\frac{0+0+(n-1) \cdot 1.5}{2}\right) \cdot T=72 \sqrt{2}$, so $K=72$, and $n=24$. Thus the desired sum is 414.

Final answer: 414