## Precalculus (6th Edition) Blitzer

First three terms : $2,8,14$ ; Last Term :$116$ and Sum $1180$
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 $1180$