## College Algebra (6th Edition)

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
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