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等于多少?

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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等于多少?
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等于多少?

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等于多少?
S(2m-1)=38=(2m-1)am
a(m-1)+a(m+1)=2am=(am)^2,am=2,
2m-1=38/am=19
m=10

10