Correct Answer - B
Here `a_(2)=13`
`rArr a+(2-1)d=13 rArr a+d=13 " " ...(1)`
and `a_(5)=25`
`rArr a+(5-1)d=25 rArr a+4d=25 " " ...(2)`
Subtracting equation (1) from equation (2), we get
`{:(""a+4d""=25),(underset(-)" "a underset(-)+d=""underset(-)""13),(bar(" "3d=12" ")):}`
`rArr " " d=4`
Put d=4 in equation (1), we get
`a+4=13 rArr a=9`
Now `a_(7)=a+(7-1)d=9+6(4)=33`