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
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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](/uploads/image/z/1348319-47-9.jpg?t=%E5%B7%B2%E7%9F%A5%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E5%85%B12n%2B1%E9%A1%B9%2C%E5%85%B6%E4%B8%AD%E6%B0%94%E6%95%B0%E9%A1%B9%E4%B9%8B%E5%92%8C%E4%B8%BA290%2C%E5%81%B6%E6%95%B0%E9%A1%B9%E4%B9%8B%E5%92%8C%E4%B8%BA261%2C%E6%B1%82%E7%AC%ACn%2B1%E9%A1%B9%E5%8F%8A%E9%A1%B9%E6%95%B02n%2B1%E7%9A%84%E5%80%BC)
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
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
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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