设{an}是等差数列,a1=19.a21=3,数列{|an|}的前n项和为sn,求sn

公差d=（a21-a1）/20=-0.8
sn=na1+n(n-1)*d/2=19n-0.4n(n-1)

公差d
d=(a21-a1)/20=-4/5
an=a1+(n-1)d>0,n<=24
a25=-0.2
1<=n<=24,an>0;n>=25,an<0
n<=24,
Sn=19n-2n(n-1)/5
n>=25,
S24=24*19+24(24-1)(-4/5)/2=235.2
Sn=235.2-(n-24)*(-0.2)-(n-24)(n-25)*d/2
=470.2+(2n^2-2399n)/5
总上：
n<=24,Sn=19n-0.4n(n-1)
n>=25,Sn=470.2+0.2(2n^2-2399n)