a1+a3+a5+...+a(2n-1)+a(2n+1)=290 (1)
a2+a4+a6+```a(2n)=261 (2)
(1)-(2)得
a(2n+1)-n*d=29 (3) (d是公差)
又 a(2n+1)=a1+(2n+1-1)*d =a1+2nd (4)
(4)带入(3)得 a1+nd=29 即 a(n+1)=29
又 S(2n+1)=[a1+a(a2n+1)]*(2n+1)/2 首项+尾项的和乘以项数除以2
= 2*a(n+1)*(2n+1)/2
= a(n+1)*(2n+1)
=29*(2n+1)=290+261=551
n=9
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已知等差数列{an}共2n+1项,其中气数项之和为290,偶数项之和为261,求第n+1项及项数2n+1的值
已知等差数列{an}共2n+1项,其中气数项之和为290,偶数项之和为261,求第n+1项及项数2n+1的值
数学人气:372 ℃时间:2019-10-29 19:10:31
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