A) 4
B) 5
C) 6
D) 7
E) 8
Correct Answer: C
Solution :
Given, \[{{S}_{2n}}=3{{S}_{n}}\] \[\frac{2n}{2}[2a+(2n-1)d]=.3\frac{n}{2}[2a+(n-1)d]\] Where, a and d are first term and common difference of an AP respectively. \[\Rightarrow \] \[4a+2(2n-1)d=6a+3(n-1)d\] \[\Rightarrow \] \[2a+(3n-3-4n+2)d=0\] \[\Rightarrow \] \[2a+(-n-1)d=0\] \[\Rightarrow \] \[2a+(n+1)(-d)=0\] \[\Rightarrow \] \[2a=(n+1)d\] ?..(i) Now, \[\frac{{{S}_{3n}}}{{{S}_{n}}}=\frac{\frac{3n}{2}[2a+(3n-1)d]}{\frac{n}{2}[2a+(n-1)d]}\] \[=\frac{3[(n+1)d+(3n-1)d]}{[(n+1)d+(n-1)d]}\][from Eq. (i)] \[=\frac{3[(n+1+3n-1)d]}{(n+1+n-1)d}=\frac{3(4nd)}{(2nd)}=6\] \[\Rightarrow \] \[{{S}_{3n}}=6{{S}_{n}}\] On compare with, \[{{S}_{3n}}=k{{S}_{n}}\] \[\Rightarrow \] \[k=6\]You need to login to perform this action.
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