Answer
First three terms : $2,8,14$ ; Last Term :$116$ and Sum $1180$
Work Step by Step
Here, $a_n=6i-4$
Plug $i=1,2,3$, then we have
$a_1=6(1)-4 =2 \\a_2=6(2)-4 =8 \\a_3=6(3)-4 =14$
Last term : $a_{20}=6(20)-4 =116$
The sum of an arithmetic sequence is given by: $S_n=\dfrac{n}{2}[a_1+a_n]$
or, $S_{20}=\dfrac{20}{2}[2+116]=1180$
Hence, First three terms : $2,8,14$ ; Last Term :$116$ and Sum $1180$
