Question

设数列{an}为等差数列，其前n项和为Sn,且S4=-62,S6=-75

求|a1|+|a2|+……+|a14|的值

Answer 1

Sn=a1n + n(n-1)d/2
S4=4a1+6d=-62
S6=6a1+15d=-75
a1=-20，d=3
an=a1+(n-1)d=3n-23
当n＜8时，an＜0
当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
= -2×(-77)-7
=147

Answer 2

s4=4a1+6d=-62,s6=6a1+15d=-75解出a1=-20,d=3
原式=-a1-a2-a3......-a7+a8+a9+....+a14=s14-2s7=147