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The sum of the first n terms of an AP is:
A$S_n = \dfrac{n}{2}[2a + (n-1)d]$
B$S_n = \dfrac{n}{2}[a + (n-1)d]$
C$S_n = \dfrac{n}{2}[2a + nd]$
D$S_n = n[2a + (n-1)d]$
Answer & Solution
Correct answer: A. $S_n = \dfrac{n}{2}[2a + (n-1)d]$
Sₙ = n/2 [2a + (n−1)d].
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