已知等差数列an满足a2+a4＝-6,a3+a5＝-2（1）求数列an的通项公式（2）求数列{｜an｜｝的前n项和Tn

(1)
an=a1+(n-1)d
a2+a4=-6
2a1+4d=-6
a1+2d=-3 (1)
a3+a5=-2
2a1+6d=-2
a1+3d=-1 (2)
(2)-(1)
d=2
a1= -7
an = -7+(n-1)2 = 2n-9
(2)
an >0
2n-9 >0
n> 9/2
|an| = -an ; n=1,2,3,4
= an ; n>=5
Tn = |a1|+|a2|+...+|an|
= -(a1+a2+a3+a4) +(a5+a6+...+an)
= (7+1)2 + ( 2n-8)(n-4)/2
= 16 + (n-4)^2
=n^2-8n + 32

（1）a2+a4=2a3=-6 a3=-3
a3+a5=2a4=-2 a4=-1
所以d=a4-a3=-1+3=2
所以an=a3+(n-3)d=-3+2(n-3)=2n-9
（2）当n<=4时，an<0
当n>=5时，an>0
所以当n<=4时，Tn=-Sn=-(-7+2n-9)*n/2=n(8-n)
当n>=5时，Tn=T4+(Sn-S4)=4*(8-4)+n(n-8)-4*(4-8)=n^2-8n+32

分别把a2=a1+d a4=a1+3d a3=a1+2d a5=a1+4d 求出a1=-7 d=2 则an=2n-9 Tn=n^2-8n (都用公式可以求出！)