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( ) A.38 B.20 C.10 D.9
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( )
A. 38
B. 20
C. 10
D. 9
A. 38
B. 20
C. 10
D. 9
数学人气:407 ℃时间:2019-10-26 07:38:59
优质解答
根据等差数列的性质可得:am-1+am+1=2am,则am-1+am+1-am2=am(2-am)=0,解得:am=0或am=2,若am等于0,显然S2m-1=(2m−1)(a1+a2m−1)2=(2m-1)am=38不成立,故有am=2,∴S2m-1=(2m-1)am=4m-2=38,解得m=10.故...
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