(1) From the sum of the first n terms in arithmetic progression and Sn=na 1+n(n+ 1)d/2, it can be seen that the sum of the first n terms in arithmetic progression has no constant term.

∴q=0

(2)a 1=S 1=p-2，a5=S5-S4=9p-2，

∫a 1+a5 = 2× 18。

∴(p-2)+(9p-2 )=36

∴ p=4

∴a 1=2，a5=34

Tolerance d=(34-2)/4=8

∴an=2+(n- 1)*8=8n-6

And an = 2 log2 bn,

∴4n-3=log2^bn

∴bn=2^(4n-3)=2* 16^(n- 1)

∴ sequence {bn} TN = 2 * (1-kloc-0/6n)/(1-kloc-0/6) = (2 *16n-2)/65438+.