Answer:
418
Sn = ∑ 3 +4(n-1)
n=1
Step-by-step explanation:
3,7,11,15,.....1671
a1 =3
d = 4 (7-3=4)
an = a1 + d(n-1)
an = 3 +4(n-1)
We need to determine the last term
1671 = 3 +4(n-1)
Subtract 3
1671-3 = 3-3 +4(n-1)
1668 = 4(n-1)
divide by 4
1668/4 = 4(n-1)/4
417 = n-1
Add 1 to each side
418 = n
418
Sn = ∑ a1 +d(n-1)
n=1
418
Sn = ∑ 3 +4(n-1)
n=1
