Answer
$4901$.
Work Step by Step
The nth term of the arithmetic sequence is given by:
$a_n=a_1+(n-1) d \\ 50=8+\dfrac{1}{4}(n-1) \\168 =(n-1) \\ n= 169$
We see that there is a constant difference between the terms of $d=\dfrac{1}{4}$ and the terms are part of an arithmetic sequence.
The terms of the sum are the first $193$ terms of an arithmetic sequence, starting with $a_{1}=8$ and with a difference of $d=\dfrac{1}{4}$.
The sum of the first $n$ terms of an arithmetic sequence is given by:
$S_{n}= \dfrac{n}{2}\left(a_{1}+a_{n}\right)$
Now, $S_{169}= \dfrac{169}{2}[8+50] \\=(169)(29) \\= 4901$
Therefore, the sum of the arithmetic sequence is: $4901$.
