等差数列{an}的前n项和为Sn,已知a(m-1)+a(m+1)-am²=0,S(2m-1)=38,则m=?
第一题:等差数列{an}的前n项和为Sn,已知a(m-1)+a(m+1)-(am)^2=0,S(2m-1)=38,则m=
一道高中等差数列题等差数列{A n}的前n项和为Sn,已知A(m-1)+A(m+1)-a(m)的平方=0,S(2m-1)
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
已知非负等差数列{an}的公差d不为0,前n项和为Sn,设m,n,p∈N*,且m+n=2p (1)求证:1/Sn+1/S
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( )
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
已知等差数列{An}前n项的和Sn,若Sm/Sn=m²/n²,则a5/a6
等差数列{an}的前几项和为Sn ,已知(m+1项)+(m-1项)-(m项的平方)=0 ,S(2m-1)=38,则m值为
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
1.等差数列{an},{bn}的前n项和分别为Sn,Tn,(1)若Sm=n,Sn=m,求Sn+m
已知等差数列{An}前n项和为Sn,且Sm/Sn=m^2/n^2,m≠n,A1=1,则An
已知等差数列{an}前n项和为Sn,若Sm/Sn=m^2/n^2,则am/am的值为?