Answer
First term(i = 1) $a_{1}$= (-2$\times$1 + 6) = -2 + 6 = 4
Second term(i = 2)$a_{2}$ = (- 2$\times$2 + 6) = -4 + 6 = 2
Third term(i = 3)$a_{3}$ = (-2$\times$3 + 6) = -6 + 6 = 0
Last term $a_{l}$(i = 40) = $a_{40}$ = (-2$\times$40 + 6) = -80 + 6 = -74
The sequence becomes = 4 + 2 + 0 +......+ (-74)
Indicated Sum = 4 + 2 + 0 +......+ (-74) = -1400
Work Step by Step
First term(i = 1) $a_{1}$= (-2$\times$1 + 6) = -2 + 6 = 4
Second term(i = 2)$a_{2}$ = (- 2$\times$2 + 6) = -4 + 6 = 2
Third term(i = 3)$a_{3}$ = (-2$\times$3 + 6) = -6 + 6 = 0
Last term $a_{l}$(i = 40) = $a_{40}$ = (-2$\times$40 + 6) = -80 + 6 = -74
The sequence becomes = 4 + 2 + 0 +......+ (-74)
Common difference (d)= 0 - 2 = 2 - 4 = -2
Number of terms (n) = 40
Indicated Sum = 4 + 2 + 0 +......+ (-74)
= $\frac{n}{2}$(first term + last term)
= $\frac{40}{2}$(4 + (-74)
= $\frac{40}{2}$(-70)
= 20$\times$ (-70)
= - 1400
