Answer
Put i = 1, 2, 3 & 50 in (-4i)
First term(i = 1) $a_{1}$= (-4$\times$1) = -4
Second term(i = 2)$a_{2}$ = (-4$\times$2) = -8
Third term(i = 3)$a_{3}$ = (-4$\times$3) = -12
Last term $a_{l}$(i = 50) = $a_{50}$ = (-4$\times$50) = -200
The sequence becomes = -4 - 8 - 12 -......- 200
Indicated Sum = -4 - 8 - 12 -.....- 200 = -5100
Work Step by Step
Put i = 1, 2, 3 & 50 in (-4i)
First term(i = 1) $a_{1}$= (-4$\times$1) = -4
Second term(i = 2)$a_{2}$ = (-4$\times$2) = -8
Third term(i = 3)$a_{3}$ = (-4$\times$3) = -12
Last term $a_{l}$(i = 50) = $a_{50}$ = (-4$\times$50) = -200
The sequence becomes = -4 - 8 - 12 -......- 200
Common difference (d)= -12 + 8 = -8 + 4 = -4
Number of terms (n) = 50
Indicated Sum = -4 - 8 - 12 -.....- 200
= $\frac{n}{2}$(first term + last term)
= $\frac{50}{2}$(-4 -200)
= 25$\times$ -204
= -5100
