a2=S2-S1=(4+2p+q)-(1+p+q)=p+3
a3=S3-S2=(9+3p+q)-(4+2p+q)=p+5
a5=S5-S4=(25+5p+q)-(16+4p+q)=p+9
已知a2,a3,a5为等比数列,则：a3²=a2*a5
===> (p+5)²=(p+3)(p+9)
===> p²+10p+25=p²+12p+27
===> p=-1
所以,a2=2,a3=4,a5=8
且,等差数列公差d=a3-a2=(p+5)-(p+3)=2
所以,a1=a2-d=0
则,an=a1+(n-1)d=2(n-1)
Sn=na1+[n(n-1)d/2]=n(n-1)=n²-n
所以,p=-1,q=0
满足an+log2bn?——这个是不是有问题?!