yhkn.net
当前位置:首页 >> 设Sn为等差数列{An}的前n项和,满足S4=14,S10%S7=... >>

设Sn为等差数列{An}的前n项和,满足S4=14,S10%S7=...

∵为等差数列{an}的前n项和,满足S4=14,S10-S7=30,∴4a1+4×32d=14a8+a9+a10=30,∴2a1+3d=7a1+8d=10,解得a1=2,d=1,∴an=2+1(n-1)=n+1,∴Sn=2n+n(n?1)2×1=n22+3n2.

S10-S7=a1+a2+a3+a4+a5+a6+a7+a8+a9+a10-( a1+a2+a3+a4+a5+a6+a7) = a8+a9+a10= 2a9+a9= 3a9=30 则a9=10 则a9=a1+8d=10 S4=4a1+6d=14 则d=1, a1=2 S9=2×9+½×8×9=54

(1)由题意可知,a3=a1+2d=3S4=4a1+4×32d=10?a1=1d=1,所以an=1+(n-1)×1=n.(2)方法1:因为公差d=1>0,所以等差数列为递增数列,所以Sn≥S1=1.方法2:Sn=n(n+1)2=12(n+12)2?18,对称轴为n=?12,所以当n=1时,Sn最小为S1=1.(3)...

∵等差数列{an}∴Sn,S2n-Sn,S3n-S2n,S4n-S3n成等差数列,又Sn=2,S2n=14,Sn=2,S2n-Sn=12,S3n-S2n=22,S4n-S3n=32,∴S4n=68.故选A.

网站首页 | 网站地图
All rights reserved Powered by www.yhkn.net
copyright ©right 2010-2021。
内容来自网络,如有侵犯请联系客服。zhit325@qq.com