Then a (n+1)-an = 2sn-2s (n-1) = 2 [sn-s (n-1)] = 2an.

So a(n+ 1)=3an, that is, a(n+ 1)/an=3, which is a constant.

So the sequence {an} is a geometric series with 1 as the first term and 3 as the common ratio.

So an = 3 (n- 1) (n ∈ n+)

(2) T3 = b1+B2+B3 = 3b2 =15, so b2=5.

Let the tolerance be d, then b 1=b2-d=5-d, B3 = B2+d = 5+d.

And a 1= 1, a2=3, a3=9, so a 1+b 1=6-d, a2+b2=8, a3+B3 =14+d.

Then according to the meaning of the question: (6-d) (14+d) = 8 2 = 64.

After sorting, (d-2)(d+ 10)=0, so d=2, or d=- 10.

And that term of the sequence {bn} is positive, so d≥0, so d=2.

So b 1=3, then bn=3+2(n- 1)=2n+ 1.

So TN = n (3+2n+ 1)/2 = n 2+2n.