Consider the arithmetic series [tex] _{t=1}\Sigma^{t=18} (3t-4) [/tex]
Let t=1 in the given series, we get
first term = [tex] a_{1} [/tex] = 3-4 = -1.
Let t=2 in the given series, we get
second term = [tex] a_{2} [/tex] = [tex] (3 \times 2)-4 = 2 [/tex]
Let t=3 in the given series, we get
third term = [tex] a_{3} = (3 \times 3)-4=5 [/tex]
Now, let t=18 in the given series, we get
last term = l = [tex] l = (3 \times 18)-4 = 50 [/tex]
We get the series as
-1, 2, 5,..... 50
Sum = [tex] \frac{n}{2}(a+l) [/tex]
= [tex] \frac{18}{2}(-1+50) [/tex]
= [tex] = 9 \times 49 [/tex]
= 441
Therefore, the sum of the given arithmetic series is 441.
