#### Answer

First term(i = 1) $a_{1}$= (6$\times$1 - 4) = 6 - 4 = 2
Second term(i = 2)$a_{2}$ = (6$\times$2 - 4) = 12 - 4 = 8
Third term(i = 3)$a_{3}$ = (6$\times$3 - 4) = 18 - 4 = 14
Last term $a_{l}$(i = 20) = $a_{20}$ = (6$\times$20 - 4) = 120 - 4 = 116
The sequence becomes = 2 + 8 + 14 +......+ 116
Indicated Sum = 2 + 8 + 14 +......+ 116 = 1180

#### Work Step by Step

First term(i = 1) $a_{1}$= (6$\times$1 - 4) = 6 - 4 = 2
Second term(i = 2)$a_{2}$ = (6$\times$2 - 4) = 12 - 4 = 8
Third term(i = 3)$a_{3}$ = (6$\times$3 - 4) = 18 - 4 = 14
Last term $a_{l}$(i = 20) = $a_{20}$ = (6$\times$20 - 4) = 120 - 4 = 116
The sequence becomes = 2 + 8 + 14 +......+ 116
Common difference (d)= 8 - 2 = 6
Number of terms (n) = 20
Indicated Sum = 2 + 8 + 14 +......+ 116
= $\frac{n}{2}$(first term + last term)
= $\frac{20}{2}$(2 + 116)
= $\frac{20}{2}$(118)
= 10 $\times$ 118
= 1180