正项数列{an}的前n项和Sn满足10Sn=an^2+5an+6,且a1,a3,a15成等比数列,则a2010=

10Sn=(an)²+5an+6
10S(n-1)=(a(n-1))²+5a(n-1）+6
两式相减,得
5a(n-1)+5an=(an)²-(a(n-1))²
5=an-a(n-1)
所以{an}是等差数列,首项a1,公差d=5,所以
an=na1+(n-1)n/2
a1*a15=(a3)²
a1*(a1+14d)=(a1+2d)²
5d*a1=2d²
d(5a1-2d)=0
∵d=5
所以5a1=2d
a1=2
∴an=a1+(n-1)d=2+(n-1)=5n-3
当n=2010时,得
a2010=10047