Match the AP's given in column A with suitable common differences given in column B.

Explanation

`A_(1).2,-2,-6,-10, …`
Here, common difference, `d= -2-2= -4`
`A_(2). :' a_(n)=a+(n-1)d`
`implies 0=-18+(10-1)d`
` " "18=9d`
`:.` Common difference, `d=2`
`A_(3). :' a_(10)=6`
`implies a+(10-1)d=6 `
` implies " " 0+9d=6 " " [:' a=0("given")]`
`implies " " 9d=6implies d=(2)/(3)`
`A_(4). :' a_(2)=13`
`implies a+(2-1)d=13 " " [ :' a_(n)=a+(n-1)d]`
` implies " " a+d=13 " " ...(i)`
` "and " a_(4)=3implies a+(4-1)d=3`
` :. " " a+3d=3 " " ...(ii) `
On subtracting Eq. (i) from Eq. (ii), we get
` " " 2d= -10`
`implies d= -5`
` :. (A_(1)) to B_(4),(A_(2)) to B_(5), (A_(3)) to B_(1)` and `(A_(4)) to B_(2)`