Question

在等差数列an中a16+ a17+a18=a9=-36，其前n项和为sn，求an的绝对值的前n项和

在等差数列an中a16+ a17+a18=a9=-36，其前n项和为sn，求an的绝对值的前n项和Tn

Answer 1

an =a1+(n-1)d
a16+a17+a18 = a9 = =-36
3a1+48d = -36 (1)
a1+8d = -36 (2)
(1)-3(2)
24d = 72
d = 3
a1= -60
an = -60+(n-1)3 = 3n - 63
Sn = (3n-123)n/2

an >0
3n-63 >0
n> 21
|an| = -an ; n≤21
= an ; n ≥22
for n≤21
|a1|+|a2|+...+|an|
= -(3n- 63 -60 ) n/2
= (123-3n)n/2
for n≥22
|a1|+|a2|+...+|an|
=S21 +a22+a23+....+an
= 630 + (3n-63)(n-20)/2

Answer 2

a16+a18=2a17
a16+ a17+a18=3a17=-36
a17=-12; a9=-36
a17=a9+8d
-12=-36+8d
d=3, a1=a9-8d=-36-8*3=-60
易得 a21=a1+20d=0,
当n≤21时 an，an-1……a1，均小于零 Tn=-Sn=-[na1+d*n（n-1）/2]=-[-60n+3n(n-1)/2]
=61.5n-1.5n^2
当n>21时，a22=3 ,T21=61.5*21-1.5*21^2=630
Tn=T21+(a22+……an)=630+[3+d(n-21)(n-22)/2]=633+1.5(n-21)(n-22)
可看成以a22为首项，3为公差的等差数列，从a22到an共有n-21项

追问

谢谢

追答

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Answer 3

a9=-36
a17 = -36/3 =-12
d=(-12 +36)/(17-9)=3
a1=a9 -8d = -60
an=a1+(n-1)d = -60+3(n-1)= 3n -63
n<21: an<0
n>=21: an>=0
{an}前n项的和Sn= [a1 + (n-1)d/2]n =(1/2) n(3n -123)
S21 =-630
设an的绝对值的前n项和为Tn
当 n>21时 Tn= Sn +2|S21| =(1/2) n(3n -123)+1260
所以 an的绝对值的前n项和为
Tn = (1/2) n(123-3n) （n<=21）
Tn= (1/2) n(3n -123)+1260 （n>21）