高三数学等差数列问题(过程)

an=3n-23,求｜an｜的前n的项和Tn
解:an=3n-23为等差数列,a1=-20,公差d=3,a7=-2
令an=3n-23≥0,得n≥23/3>7,故{an}从第8项起为正,前7项为负
记{an}前n的项和Sn
1.当n≤7
Tn=-Sn=-[n(a1+an)/2]=-[n(-20+3n-23)/2]=-[n(3n-43)/2]=[n(43-3n)]/2
2.当n>7
Tn=(-a1-a2...-a7)+(a8+...+an)
=(-S7)+(Sn-S7)
=Sn-2S7
=[n(3n-43)/2]-2*7(a1+a7)/2
=[n(3n-43)/2]-7(-20-2)
=[n(3n-43)/2]+154