设等差数列{an}的前n项和为Sn,Sm-1=-2,Sm=0,Sm 1=3,则m=
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an =a1+(n-1)d
Sn= (2a1+(n-1)d)n/2
S(m-1) =(2a1+(m-2)d)(m-1)/2 = -2 (1)
Sm = (2a1+(m-1)d)m/2 =0 (2)
S(m+1) =(2a1+md)(m+1)/2=3 (3)
(2)-(1)
md(m-1)- (m-1)(m-2)d = 4
md -d=2
d= 2/(m-1)
(3)-(2)
md(m+1)- (m-1)md = 6
md=3
2m/(m-1) = 3
2m=3m-3
m=3
Sn= (2a1+(n-1)d)n/2
S(m-1) =(2a1+(m-2)d)(m-1)/2 = -2 (1)
Sm = (2a1+(m-1)d)m/2 =0 (2)
S(m+1) =(2a1+md)(m+1)/2=3 (3)
(2)-(1)
md(m-1)- (m-1)(m-2)d = 4
md -d=2
d= 2/(m-1)
(3)-(2)
md(m+1)- (m-1)md = 6
md=3
2m/(m-1) = 3
2m=3m-3
m=3
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