Answer
Arithmetic
Sum: $-\dfrac{2225}{4}$
Work Step by Step
We are given the sequence:
$\left\{8-\dfrac{3}{4}n\right\}$
Compute the difference between two consecutive terms:
$\left(8-\dfrac{3}{4}(n+1)\right)-\left(8-\dfrac{3}{4}n\right)=8-\dfrac{3}{4}n-\dfrac{3}{4}-8+\dfrac{3}{4}n=-\dfrac{3}{4}$
As the difference between any consecutive terms is constant, the sequence is ARITHMETIC. Determine its first element and common difference:
$a_1=8-\dfrac{3}{4}(1)=\dfrac{29}{4}$
$d=-\dfrac{3}{4}$
Compute the sum of the first 50 terms:
$S_n=\dfrac{n(2a_1+(n-1)d)}{2}$
$S_{50}=\dfrac{50\left(2\left(\dfrac{29}{4}\right)+(50-1)\left(-\dfrac{3}{4}\right)\right)}{2}=-\dfrac{2225}{4}$
