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

Explanation

Let the successive terms of an AP be `t_(1),t_(2),t_(3),"....,"t_(9),t_(10)`.
By hypothesis,
`t_(2),t_(4),t_(6),t_(8),t_(10)=15`
`implies (5)/(2)(t_(2)+t_(10))=15`
`implies t_(2)+t_(10)+6`
`implies (a+d)+(a+9d)=6`
`implies 2a+10d=6 " " "......(i)"`
and `t_(1)+t_(3)+t_(5)+t_(7)+t_(9)=12(1)/(2)`
`implies (5)/(2)(t_(1)+t_(9))=(25)/(2)`
`implies t_(1)+t_(9)=5`
`implies a+a+8d=5`
`implies 2a+8d=5 " " ".....(ii)"`
From Eqs. (i) and (ii), we get `d=(1)/(2) " and " a=(1)/(2)`
Hence, the AP is `(1)/(2),1,1(1)/(2),2,2(1)/(2),"...."`