等差数列{an}的前n项和为Sn,已知an-1+an+1-an2=0,S2n-1=38,则n=( )
A.38
B.20
C.10
D.9
因为an是等差数列,所以an-1+an+1=2an,由an-1+an+1-an2=0,
得:2an-an2=0,所以an=2,又S2n-1=38,即
=38,(2n-1)(a1+a2n-1) 2
即
=38(2n-1)•2an 2
即(2n-1)×2=38,解得n=10.
故选C.
等差数列{an}的前n项和为Sn,已知an-1+an+1-an2=0,S2n-1=38,则n=( )
A.38
B.20
C.10
D.9
因为an是等差数列,所以an-1+an+1=2an,由an-1+an+1-an2=0,
得:2an-an2=0,所以an=2,又S2n-1=38,即
=38,(2n-1)(a1+a2n-1) 2
即
=38(2n-1)•2an 2
即(2n-1)×2=38,解得n=10.
故选C.