If the arithmetic progression whose common difference is nonzero the sum of first `3n` terms is equal to the sum of next `n` terms. Then, find the ratio of the sum of the `2n` terms to the sum of next `2n` terms.

Explanation

Given `S_(3n)=S_(n)'=s_(4n)-S_(3n)`
or `2S_(3n)=S_(4n)`
or `2(3n)/2(2a+(3n-1)d)=(4n)2(2a+(4n-1)d)`
or 12a+(18n-6)d=8a+(16n-4)d
or 4a=(-2n+2)d
or 2a=(1-n)d
Now we have to find `(S_(2n))/(S_(2n'))`
`(S_(2n))/(S_(2n)')=(S_(2n))/(S_(4n)-S_(2n))`
`=((2n)/2(2a+(2n-1)d))/((4n)/2[2a+(4n-1)d]-(2n)/2[2a+(2n-1)d])`
`(2[(1-n)d+(2n-1)d])/(4[(1-n)d+(4n-1)d]-2[(1-n)d+(2n-1 )d])`
`=(2nd)/(10nd)=1/5`