解:∵10Sn=an2+5an+6, ①
∴10a1=a12+5a1+6,解之得a1=2或a1=3.
又10Sn-1=an-12+5an-1+6(n≥2), ②
由①-②得10an=(an2-an-12)+6(an-an-1),
即(an+an-1)(an-an-1-5)=0.
∵an+an-1＞0,∴an-an-1=5(n≥2).
当a1=3时,a3=13,a15=73.
a1,a3,a15不成等比数列,∴a1≠3.
当a1=2时,a3=12,a15=72,有a32=a1a15.
∴a1=2.
∴an=5n-3.