等差数列中,公差d≠0,|a11|=|a51|,a20=22,设{an}前n项和为Sn,{|an|}前n项和为Tn.(1

d≠0,|a11|=|a51|
-a11 = a51
-(a1 + 10d) = a1 + 50d
a1 = -30d
a20 = a1 + 19d
-30d + 19d = 22
d = -2
a1 = 60
an = a1 + (n-1)d = 60 + (n-1)*(-2) = 62 - 2n
|an| = |62 -2n|
通项公式可以加绝对值符号
|an| = | 62 - 2n|
当 n ≤ 31 时,
|an| = 62 - 2n
当 n > 31 时
|an| = 2n -62
因此 当 n ≤ 31 时
Tn = (|a1| + |an|)*n/2
= (60 + 62 -2n)*n/2
= n(61 -n)
T31 = 31(61 -31) = 930
当 n > 31 时
Tn = T31 + (|a31| + |an|)*(n-30)/2
这里就是把 项数 变成 n-30 了
其中 a31 = 62 - 2*31 = 0
Tn = T31 + (0 + 2n -62)(n-30)/2
= T31 + (n-31)(n-30)
= 930 + (n-31)(n-30)
= n^2 - 61n + 1860

公差d≠0,|a11|=|a51|,说明a11+a51=0
a11+a51=2a31=0
而a20=22>0
a31-a20=11d<0
d=-2<0
a1=a31-30d=60
an=a1+(n-1)d=62-2n
(1)
当n<=31时
|an|=62-2n
当n>31时
|an|=2n-62
(2...