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If the sum of the first $n$ terms of a sequence is $S_n = n^2$, its $n$th term $a_n$ is
A$2n - 1$
B$2n + 1$
C$n^2 - 1$
D$2n$
Answer & Solution
Correct answer: A. $2n - 1$
1. Use $a_n = S_n - S_{n-1}$.
2. $a_n = n^2 - (n-1)^2 = n^2 - (n^2 - 2n + 1) = 2n - 1$.
3. So $a_n = 2n - 1$ (the sequence of odd numbers, whose partial sums are $n^2$).
_Source: Karnataka SSLC (KSEEB) Class 10 Mathematics, Ch1 'Arithmetic Progressions' ($a_n = S_n - S_{n-1}$)_
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