When the ninth term of an AP is divided by its second term we get 5 as the quotient, when the thirteenth term ia devided ny sixth term the quotient is 2 and the remainderis 5, then the seonnd term is

Explanation

Let a be the first term ad d be the common difference, then`T_(9)=5T_(2)`
`implies (a+8d)=5(a+d)`
`:. 4a=3d" " "….(i)"`
and `T_(13)=T_(6)xx2+5`
`implies a+12d=2(a+5d)+5`
`implies 2d=a+5" " "….(ii)"`
From Eqs.(i) and (ii), we get
`a=3` and `d=4`
`:.T_(2)=a+d=7`