A) \[2{{S}_{n}}\]
B) \[{{S}_{3n}}\]
C) \[\frac{1}{3}{{S}_{3n}}\]
D) \[\frac{1}{2}{{S}_{n}}\]
Correct Answer: C
Solution :
\[{{S}_{2n}}-{{S}_{n}}=\frac{2n}{2}\{2a+(2n-1)d\}-\frac{n}{2}\{2a+(n-1)d\}\] \[=\frac{n}{2}\{4a+4nd-2d-2a-nd+d\}=\frac{n}{2}\{2a+(3n-1)d\}\] \[=\frac{1}{3}.\frac{3n}{2}\{2a+(3n-1)d\}=\frac{1}{3}{{S}_{3n}}\].
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