Answer
$1120$
Work Step by Step
There are $80$ terms, and they form an arithmetic sequence with
$d=a_{n+1}-a_{n}=\displaystyle \frac{1}{3}(n+1)+\frac{1}{2}-(\frac{1}{3}n+\frac{1}{2})=\frac{1}{3}$
$a_{1}=\displaystyle \frac{1}{3}(1)+\frac{1}{2}=\frac{5}{6},$
$a_{80}=\displaystyle \frac{1}{3}(80)+\frac{1}{2}=\frac{163}{6}$
Sum of the First $n$ Terms of an arithmetic sequence:
$S_{n}=\displaystyle \frac{n}{2}\left(a_{1}+a_{n}\right)$
$S_{80}=\displaystyle \frac{80}{2}\left(\frac{5}{6}+\frac{163}{6}\right)$
$=40(\displaystyle \frac{168}{6})=40(28)=1120$
