## Precalculus (10th Edition)

$-1925$
We have to determine the sum: $S=\sum_{n=1}^{100} \left(6-\dfrac{1}{2}n\right)$ $S$ is the sum of an arithmetic sequence. Determine its first term and the common difference: $a_1=6-\dfrac{1}{2}(1)=\dfrac{11}{2}$ $d=\left(6-\dfrac{1}{2}(k+1)\right)-\left(6-\dfrac{1}{2}k\right)$ $=6-\dfrac{1}{2}(k+1)-6+\dfrac{1}{2}k=-\dfrac{1}{2}$ The number of terms is 100, so we have to determine the sum of the first $100$ terms. We use the formula: $S_n=\dfrac{n(a_1+a_n)}{2}$ $a_1=\dfrac{11}{2}$ $a_{100}=6-\dfrac{1}{2}(100)=-44$ $\sum_{n=1}^{100} \left(6-\dfrac{1}{2}n\right)=\dfrac{100\left(\dfrac{11}{2}-44\right)}{2}=-1925$