由a 6,a 9,a 15依次为等比数列得到a 9 2=a 6a 15即(a 1+8d) 2=(a 1+5d)(a 1+14d),
化简得3d(a 1+2d)=0,由d≠0,得到a 1=-2d,
所以数列{b n}的公比q=
a 9
a 6 =
-2d+8d
-2d+5d =2,首项b 1=
1
2 ,
则S 5=
1
2 (1- 2 5 )
1-2 =
31
2
故答案为:
31
2
由a 6,a 9,a 15依次为等比数列得到a 9 2=a 6a 15即(a 1+8d) 2=(a 1+5d)(a 1+14d),
化简得3d(a 1+2d)=0,由d≠0,得到a 1=-2d,
所以数列{b n}的公比q=
a 9
a 6 =
-2d+8d
-2d+5d =2,首项b 1=
1
2 ,
则S 5=
1
2 (1- 2 5 )
1-2 =
31
2
故答案为:
31
2