Because A(n+ 1)-A(n)= 1/2, A(n) is a arithmetic progression with a first term of A 1=2 and a tolerance of 1/2.

So a (n) = a1+(n-1) *1/2 = 2+(n-1)/2.

So a101= 2+(101-1)/2 = 2+50 = 52.

(2) Because A+ 1, A+3, B and A+B become arithmetic progression; Tolerance d=(A+3)-(A+ 1)=2.

So: (A+3)+2=B, B+2 = A+B.

Solution: A=2, B=7.

(3)A 1+A 1 is not necessarily equal to A2; When this series is a constant series, it holds, and generally arithmetic progression does not hold;