(1)Sn=2an+1
Sn-1=2an-1+1
所以an=2an-2an-1
an=2an-1
故{an}是首項(xiàng)為a1=-1,公比為2的等差數(shù)列an=-2^(n-1)
(2)設(shè)an=a+(n-1)d,bn=bq^(n-1)
a5=a+4d=10
因?yàn)閧bn}是等比數(shù)列b3^2=b1*b5,所以a7^2=a5a10
即(a+6d)^2=(a+4d)(a+9d)
解方程a=0,d=5/2,a20=a+19d=95/2
b1=b=a5=10,b3=bq^2=a7=15,b5=a10=45/2=bq^4
解得b=10,q=√6/2或-√6/2,
所以bn=10*(√6/2)^(n-1)或bn=10*(-√6/2)^(n-1)