question_answer

If the arithmetic progression whose common difference is non-zero, the sum of first 3 n terms is equal to the next n terms. Then, the ratio of the sum of the first 2n terms to the next 2n terms is

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 \[\frac{{{S}_{2n}}}{S{{}_{2n}}}\] \[\frac{{{S}_{2n}}}{S{{}_{2n}}}=\frac{{{S}_{2n}}}{{{S}_{4n}}-{{S}_{2n}}}\] \[=\frac{\frac{2n}{2}[2a+(2n-1)d]}{\frac{4n}{2}[2a+(4n-1)d]-\frac{2n}{2}[2a+(2n-1)]d}\]\[=\frac{2[(1-n)d+(2n-1)d]}{4(1-n)d+(4n-1)d-2(1-n)d+(2n-1)d}\] \[=\frac{2nd}{10nd}=\frac{1}{5}\]