已知数列{an}中,a3=2,a7=1,若{1/(an+1)}为等差数列,则a11=?
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![已知数列{an}中,a3=2,a7=1,若{1/(an+1)}为等差数列,则a11=?](/uploads/image/z/8894915-35-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%AD%2Ca3%3D2%2Ca7%3D1%2C%E8%8B%A5%7B1%2F%28an%2B1%29%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%88%99a11%3D%3F)
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已知数列{an}中,a3=2,a7=1,若{1/(an+1)}为等差数列,则a11=?
已知数列{an}中,a3=2,a7=1,若{1/(an+1)}为等差数列,则a11=?
已知数列{an}中,a3=2,a7=1,若{1/(an+1)}为等差数列,则a11=?
a3=2,a7=1,若{1/(an+1)}为等差数列
公差d=(1/a7+1 -1/a3+1)/(7-3) =1/24
1/(a11 +1) =1/3 +1/24 *(11-3) =2/3
a11=1/2
可设bn=1/(1+an) n=1,2,3,..
易知, bn是等差数列
b3=1/3=b1+2d
b7=1/2=b1+6d
解得d=1/24 b1=1/4
∴bn=(n+5)/24
b11=16/24=2/3
∴2/3=1/(1+a11)
∴a11=1/2
设数列 {1an+1}的公差为d
∵数列{an}中, a3=2,a7=1,数列{1an+1}是等差数列
∴ 1a7+1=1a3+1+4d
将a3=2,a7=1代入得:d= 124
∵ 1a11+1=1a7+1+4d
∴a11= 12
1/2-1/3=1/6
1/2+1/6=1/(a11+1)=2/3
a11=1/2