∵{an}是等差数列,且a1+a2+a3=15,∴a2=5.又∵a1?a2?a3=105,∴a1a3=21.由 a1a3=21 a1+a3=10 及{an}是递减数列,可求得a1=7,d=-2.∴an=9-2n,由an≥0得n≤4,∴选A.