(1)由题意知5S 2=4S 4
∴
5 a 1 (1- q 2 )
1-q =
4 a 1 (1- q 4 )
1-q
∵a 1≠0,q>0且q≠1∴(1+q 2)=5,
∴得 q=
1
2 ;
(2)∵ S n =
a 1 (1- q n )
1-q =2 a 1 - a 1 (
1
2 ) n-1
∴ b n =q+ s n =
1
2 +2 a 1 - a 1 (
1
2 ) n-1
要使{b n}为等比数列,当且仅当
1
2 +2 a 1 =0
即 a 1 =-
1
4 ,此 b n =(
1
2 ) n+1 为等比数列,
∴{b n}能为等比数列,此时 a 1 =-
1
4 .