Answer
$4901$
Work Step by Step
There is a constant difference, $\displaystyle \frac{1}{4}$, between terms.
The terms are part of an arithmetic sequence.
nth Term of an Arithmetic Sequence:
$a_{n}=a_{1}+(n-1)d$
$ 50=8+\displaystyle \frac{1}{4}(n-1)\qquad$... solve for n
$42=\displaystyle \frac{1}{4}(n-1)$
$168=(n-1)$
$n=169$
There are $169$ terms in the sum.
The terms of the sum
are the first $169$ terms of an arithmetic sequence, $a_{1}=8, d=\displaystyle \frac{1}{4}.$
Sum of the First $n$ Terms of an arithmetic sequence:
$S_{n}=\displaystyle \frac{169}{2}\left(8+50 \right)=169\cdot 29=4901$
