Answer
$-1925$
Work Step by Step
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:
$S_n=\dfrac{n(a_1+a_n)}{2}$
$a_1=\dfrac{11}{2}$
$a_{100}=6-\dfrac{1}{2}(100)=-44$
$\sum_{n=1}^{100} \left(6-\dfrac{1}{2}n\right)=\dfrac{100\left(\dfrac{11}{2}-44\right)}{2}=-1925$
