Answer
$324$
Work Step by Step
There is a constant difference, $4$, between terms.
The terms are part of an arithmetic sequence.
nth Term of an Arithmetic Sequence:
$a_{n}=a_{1}+(n-1)d$
$49=5+4(n-1)\quad$ ... solve for n
$44=4(n-1)$
$11=(n-1)$
$n=12$
There are 12 terms in the sum.
The terms of the sum
are the first $12$ terms of an arithmetic sequence, $a_{1}=5, d=4.$
Sum of the First $n$ Terms of an arithmetic sequence:
$S_{n}=\displaystyle \frac{n}{2}\left(a_{1}+a_{n}\right)$
$S_{12}=\displaystyle \frac{12}{2}\left(5+49\right)=6\cdot 54=324$
