E
Editor
Sep 18, 2022
If a is the first term, d is common difference, then the sum of nth term of the series $S_{n}$=
Difficulty: Easy
A:
n/a[a+(n-1)d]
B:
a+(n-1)d
C:
n/2[2a+(n-1)d]
D:
n/a[2a+(n-1)d]
ID: 63273dbaca83163ac3a129a5