A) 4
B) 6
C) \[\pm 3\]
D) \[\pm 4\]
Correct Answer: B
Solution :
Given, \[{{S}_{2n}}=3{{S}_{n}}\] \[\Rightarrow \]\[\frac{2n}{2}[2a+(2n-1)d]=3.\frac{n}{2}[2a+(n-1)d]\] where a is the first term and d is the common difference of given AP. \[\Rightarrow \] \[2a=(n+1)d\] \[\therefore \] \[\frac{{{S}_{3n}}}{{{S}_{n}}}=\frac{\frac{3n}{2}[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]}\] \[=\frac{12nd}{2nd}=6\]
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