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. |