Answer
$-7592$.
Work Step by Step
The nth term of the arithmetic sequence is given by:
$a_n=a_1+(n-1) d \\ -299=7-6(n-1) \\ 51 =(n-1) \\ n= 52$
We see that there is a constant difference between the terms of $d=-6$ and the terms are part of an arithmetic sequence.
The terms of the sum are the first $52$ terms of an arithmetic sequence, starting with $a_{1}=7$ and with a difference of $d= -6$.
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, $S_{52}= \dfrac{52}{2}[7-299] \\=(26)(-292) \\=-7592$
Therefore, the sum of the arithmetic sequence is: $-7592$.
