设数列{An}的前n项和为Sn,已知A1=1,A2=6,A3=11,且（5n-8）S（n+1）—（5n+2）Sn=Pn+

（5n-8）S（n+1）—（5n+2）Sn=Pn+B
(5n-8)a(n+1)-10Sn=Pn+B (1) S（n+1）—Sn=a（n+1）
(5n-3)a(n+2)-10S(n+1)==P(n+1)+B (2) 让n变成n+1
(2)-(1)
(5n-3)a(n+2)-(5n+2)a(n+1)=P (3)
(5n+2)a(n+3)-(5n+7)a(n+2)=P (4) 让n变成n+1
(4)-(3)
(5n+2)a(n+3)+(5n+2)a(n+1)-(10n+4)a(n+2)=0
a(n+3)+a(n+1)=2a(n+2)
所以得到它是等差数列