A) \[\frac{1}{5}\]
B) \[\frac{2}{3}\]
C) \[\frac{3}{4}\]
D) None of these
Correct Answer: A
Solution :
Given, \[{{S}_{3n}}=S{{}_{n}}={{S}_{4n}}-{{S}_{3n}}\] \[\Rightarrow \] \[2{{S}_{3n}}={{S}_{4n}}\] \[\Rightarrow \]\[2.\frac{3n}{2}[2a+(3n-1)d]=\frac{4n}{2}[2a+(4n-1)d]\] \[\Rightarrow \]\[12a+(18n-6)d=8a+(16n-4)d\] \[\Rightarrow \] \[4a=(-2n+2)d\] \[\Rightarrow \] \[2a=(1-n)d\] ?(1) Now, we have to find \[\frac{{{S}_{2n}}}{S{{}_{2n}}}\] \[\frac{{{S}_{2n}}}{S{{}_{2n}}}=\frac{{{S}_{2n}}}{{{S}_{4n}}-{{S}_{2n}}}\] \[=\frac{\frac{2n}{2}[2a+(2n-1)d]}{\frac{4n}{2}[2a+(4n-1)d]-\frac{2n}{2}[2a+(2n-1)]d}\]\[=\frac{2[(1-n)d+(2n-1)d]}{4(1-n)d+(4n-1)d-2(1-n)d+(2n-1)d}\] \[=\frac{2nd}{10nd}=\frac{1}{5}\]You need to login to perform this action.
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