A) 4
B) 6
C) 8
D) 10
Correct Answer: B
Solution :
Given, sum of n terms of an\[AP={{S}_{n}}\] and \[{{S}_{2n}}=3{{S}_{n}}\] \[\Rightarrow \] \[\frac{1}{2}(2n)[2a+(2n-1)d]\] \[=3\left( \frac{n}{2} \right)[2a+(n-1)d]\] \[\Rightarrow \]\[2n[2a+(2n-1)d]=3n[2a+(n-1)d]\] \[\Rightarrow \]\[4a+2(2n-1)d=6a+3(n-1)d\] \[\Rightarrow \] \[2a=4nd-2d-3nd+3d\] \[\Rightarrow \] \[2a=nd+d\] \[\Rightarrow \] \[2a=(n+1)d\] ...(i) Now, \[\frac{{{S}_{3n}}}{{{S}_{n}}}=\frac{\frac{1}{2}(3n)[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]}\] \[=\frac{3[dn+d+3nd-d]}{nd+d+nd-d}\][using Eq. (i)] \[=\frac{12nd}{2nd}=6\]
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