Answer
$1120$.
Work Step by Step
We can see that there are $80$ terms and these terms are part of an arithmetic sequence, so we have:
$a_1=(\dfrac{1}{3})(1)+\dfrac{1}{2}=\dfrac{5}{6} \\ a_{80}=(\dfrac{1}{3})(80)+\dfrac{1}{2}=\dfrac{163}{6}$
There is a constant difference between the terms of:
$d=a_{n+1}-a_n=(\dfrac{1}{3})(n+1)+\dfrac{1}{2}-(\dfrac{1}{3}n+\dfrac{1}{2})=\dfrac{1}{3}$
The sum of the first $n$ terms of an arithmetic sequence is given by:
$S_{n}= \dfrac{n}{2}\left(a_{1}+a_{n}\right) ..(1)$
Now, we plug in the above data into Equation-1 to obtain:
$S_{80}= \dfrac{80}{2}[\dfrac{5}{6}+\dfrac{163}{6}] \\=(40)(28) \\= 1120$
Therefore, the sum of the arithmetic sequence is: $1120$.
