已知在等差数列an中1.已知d=3 an=20 sn=65 求n2.已知a11=-1 求s213.已知an=11-3n

1 因为,an=a1+(n-1)d
所以,a1=an-(n-1)d=23-3n
又,sn=a1*n+[dn(n-1)]/2
所以,sn=n*(23-3n)+[dn(n-1)]/2
sn=65,d=3
化简上式可得：n^2-3n-28=0
(n-7)(n+4)=0
所以：n=7,n=-4
n为正整数,所以n=7
2.S21
=(a1+a21)*21/2
=2a11*21/2
=21a11
=21*(-1)
=-21
3,an=11-3n
a1=11-3*1=8
a21=11-3*21=-52
s21
=(a1+a21)*21/2
=(8-52)*21/2
=-22*21
=-462