Question

等差数列{an}前n项和为Sn,且S4=2S2+4,若Sn大于等于S8,对n小于等于N*恒成立,则首项a1的范围为

Answer 1

很高兴为你解答~

S4=S2+a3+a4=2S2+4

S2+4=a3+a4

a1+a2+4=a3+a4

4d=4

d=1

由题意可知

S7≥S8,S9≧S8

7a1+7×6×1/2×1≧8a1+8×7×1/2×1

a1≤-7

9a1+9×8×1/2×1≥8a1+8×7×1/2×1

a1≥-8

综上a1∈[-8,-7]

答案无误~抄袭自觉退散!