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The sum of the first $n$ terms of an AP is given by
A$S_n = \tfrac{n}{2}[a + (n-1)d]$
B$S_n = \tfrac{n}{2}[2a + (n-1)d]$
C$S_n = n[2a + (n-1)d]$
D$S_n = \tfrac{n}{2}[2a + nd]$
Answer & Solution
Correct answer: B. $S_n = \tfrac{n}{2}[2a + (n-1)d]$
1. The sum of the first n terms is $S_n = \tfrac{n}{2}[\,2a + (n-1)d\,]$.
2. Equivalently $S_n = \tfrac{n}{2}(a + l)$ where l is the last term.
3. Both give the same value once l = a + (n−1)d is substituted.
_Source: Karnataka SSLC (KSEEB) Class 10 Mathematics, Ch1 'Arithmetic Progressions' (sum of n terms)_
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