(1)设an的公差为d,bn的公比为q,则依题意有q>0且
1+d+q2=7
1+2d+q=7
解得d=2,q=2.(2分)
所以an=1+(n-1)d=2n-1,bn=qn-1=2n-1.(4分)
(2)因为cn=an-2010=2n-2011≥0⇔n≥1005.5,
所以,当1≤n≤1005时,cn<0,当n≥1006时,cn>0.(6分)
所以当n=1005时,An取得最小值.(7分)
(3)
an
bn=
2n−1
2n−1.Sn=1+
3
21+
5
22++
2n−3
2n−2+
2n−1
2n−1①(9分)2Sn=2+3+
5
2++
2n−3
2n−3+
2n−1
2n−2②
②-①得Sn=2+2+
2
2+
2
22++
2
2n−2−
2n−1
2n−1=2+2×
1−
1
2n−1
1−
1
2−
2n−1
2n−1=6−
2n+3
2n−1.(12分)