Here, a1=a+b,a2=(a+1)+b,a3=(a+1)+(b+1)
a2−a1=(a+1)+b−(a+b)=a+1+b−a=1
a3−a2=(a+1)+(b+1)−[(a+1)+b]
=a+1+b+1−a−1−b=1
∵ a2−a1=a3−a2=1=Common difference
Since the difference between successive terms are same, it forms an AP.
The next three terms are,
a4=a1+3d=a+b+3(1)=(a+2)+(b+1)
a5=a1+4d=a+b+4(1)=(a+2)+(b+2)
a6=a1+5d=a+b+5(1)=(a+3)+(b+2)