First term (a=11). Let number of terms be (n), common difference (d).

Sum of first 4 terms: (\dfrac{4}{2}[2\cdot11+3d]=56\implies2(22+3d)=56\implies22+3d=28\implies d=2)

Last four terms: (a_{n-3}+a_{n-2}+a_{n-1}+a_n=112)

[ 4a+(4n-10)d=112 ] [ 4\cdot11+(4n-10)\cdot2=112 ] [ 44+8n-20=112 ] [ 8n+24=112\implies8n=88\implies n=11 ]

Answer: 11 terms