The correct option is B 60
First term in this series will be 12 and the last term, 248.
The the series will be 12,16,20……….248.
So, a=12, common difference = 4 and last term = 248.
so 248=12+(n-1)x4..........(assume that it has n such numbers.)
(n-1).4= 248–12=236
n-1=236/4=59
n=59+1=60.
60 multiples of 4 lie between 10 and 250.