等差数列题等差数列an的前n项和为Sn,已知am-1 + am+1 -am^2 =0,S2m-1=38,则m=

来源：学生作业帮助网 编辑：作业帮 时间：2024/05/25 16:30:45
等差数列题等差数列an的前n项和为Sn,已知am-1 + am+1 -am^2 =0,S2m-1=38,则m=
x){uϦnx1ļZv\餞';v<ݾ چ qF :F@Q[c 3smmM ;Z{$e R R"JA%B9@yC{:4!* FkHM"4SJ@C dhT,\ !-#`gZٹ6yvаssvlxkN^d*$U(|

等差数列题等差数列an的前n项和为Sn,已知am-1 + am+1 -am^2 =0,S2m-1=38,则m=
等差数列题
等差数列an的前n项和为Sn,已知am-1 + am+1 -am^2 =0,S2m-1=38,则m=

等差数列题等差数列an的前n项和为Sn,已知am-1 + am+1 -am^2 =0,S2m-1=38,则m=
am^2=am-1+am+1
am^2=am-d+am+d
am^2=2am
am=2=a1+（m-1)d
S2m-1=(2m-1)a1+(2m-1)(2m-2)d/2
=(2m-1)a1+(2m-1)(m-1)d
=(2m-1)[a1+(m-1)d]
=(2m-1)am
=(2m-1)*2=38
2m-1=19
m=10

n?m到底是哪个

设Sn为等差数列an的前n项和.求证Sn/n为等差数列 设Sn为等差数列{An}的前n项和,求证：{Sn/n}是等差数列 等差数列{an}的前n项和为sn,a10 一道关于等差数列的题设Sn为等差数列｛An｝的前n项和 求证：数列｛n分之Sn}是等差数列 (1)已知数列an的前n项和为sn满足sn=an²+bn,求证an是等差数列(2)已知等差数列an的前n项和为sn,求证数列sn/n也成等差数列 设等差数列{an}的前n项和为Sn 若a1=Sn> 等差数列{an}前n项和Sn=-n²+n,求等差数列的首项 记数列{an}的前n项和为Sn,若{Sn/an}是公差为d的等差数列,则{an}为等差数列的充要条件是d=? 已知等差数列an中,前n项和sn=n^2-15n,则使sn为最小值的n 已知公差不为0的等差数列{An}的首项A1=1,前n项和为Sn,若数列{Sn/An}是等差数列,求An? 等差数列{an}的前n项和为Sn,证明S10,S20-S10,S30-S20也成等差数列? 等差数列an的前n项和为Sn,已知a5=11 a8=5求an和Sn sn为等差数列{an}的前n项的和,已知s15>0,s16 设等差数列an的前n项和为Sn,若S4>=10,S5 等差数列{An}的前n项和为Sn,S10=45,a4+a6= 设等差数列{AN}的前N项和为SN,若S9>0,S10 设等差数列 {an}的前n 项和为Sn,若S9>0 ,S10 设等差数列{an}的前n项和为Sn,已知a3=11,S15*S16