已知Sm2 S2m-Sm S3m-S2m成等差数列 为什么(S2m-Sm)=Sm+(S3m-S2m)
已知{an}是等差数列,前n项和记为Sn,已知数列Sm,求证:Sm,S2m-Sm,S3m-S2m也成等差数列
在等差数列{an}中,公差为d,顺次m项和Sm,S2m-Sm,S3m-S2m,.组成公差为______的等差数列
已知等比数列an中,Sm=10 S2m=30
数列.已知等差数列{An}中,An≠0,若m>1且Am-1 - Am ^2 + Am+1=0,S2m-1 =38,则m=
等差数列{an}的前n项和为Sn,已知am-1 +am+1 -(am)^2=0,S2m-1=38,求m
等差数列{an}的前n项和为Sn,已知am-1+am+1-am2=0,S2m-1=38,则m等于
等差数列an的前N项和为Sn,已知am-1+am+1-am的平方=0,s2m-1=38
等差数列{An}前n项和是Sn.已知Am_1+Am+1-Am的平方等于零.S2m-1等于38.则m等于
SM,
已知等差数列{an}中的前n项和为Sn,若m>1,且am-1+am+1-am²=0,S2m-1=38,则m等于
已知等差数列{an}的前n项和为Sn,若m>1,且am-1+am+1-am2=0,S2m-1=38,则m等于( )
已知等差数列{An}的前n项和为Sn,诺m>1,且Am-1+Am+1=(Am)^2,S2m-1=38,则m为多少