A) 4
B) 6
C) 8
D) 10
Correct Answer: B
Solution :
Let the first term and common difference of the AP are 'a' and ?b?respectively, then \[{{S}_{2n}}=3{{S}_{n}}\] \[\frac{2n}{2}[2a+(2n-1)d]=\frac{3n}{2}[2a+(n-1)d]\] \[\Rightarrow \]\[4an+4{{n}^{2}}d-2nd=6an+3{{n}^{2}}d-3nd\] \[\Rightarrow \] \[2an-{{n}^{2}}d-nd=0\] \[\Rightarrow \] \[2a-nd-d=0\] \[\Rightarrow \] \[2a=nd+d\] ...(i) Now, \[\frac{{{S}_{3n}}}{{{S}_{n}}}=\frac{\frac{3n}{2}[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]}\] \[=\frac{3[2a+3nd-d]}{[2a+nd-d]}\] \[=\frac{3(nd+d+3nd-d)}{(nd+d+nd-d)}\] [Using Eq.(i)] \[=\frac{3.4nd}{2nd}=6\]
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