Question

数列｛an｝是等差数列，其前n项和是Sn,且S4＝-62,S6＝-75,｜a1｜＋｜a2｜＋...＋｜a14｜的值

要详细过程！急

Answer 1

S4=4a1+6d=-62
S6=6a1+15d=-75
a1=-20，d=3
an=a1+(n-1)d=3n-23
n＜8时,an＜0
n≥8时,an＞0
∴|a1|+|a2|+…+|a14|
= -(a1+a2+a3+a4+a5+a6+a7) +a8+a9+a10+a11+a12+a13+a14
= -S7+(S14-S7)
=-2S7+S14
=147

Answer 2

S4=4a1+6d=-62
S6=6a1+15d=-75
2a1+3d=-31 (1)
2a1+5d=-25 (2)
解得d=3 a1=-20
|a1|=20 |a2|=17 … |a7|=2
|a8|=1 |a9|=4 … |a14|=19
|a1|+|a2|+...+|a14|=2+3(0+1+...+6)+1+3(0+1+...+6)=129