搜搜做题作业网设{an}是各项互不相等的正数等差数列，{bn}是各项互不相等的正数等比数列，a1=b1，a2n+1=b2n+1，则（
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设{an}是各项互不相等的正数等差数列，{bn}是各项互不相等的正数等比数列，a1=b1，a2n+1=b2n+1，则（ ）
A. an+1＞bn+1
B. an+1≥bn+1
C. an+1＜bn+1
D. an+1=bn+1
A. an+1＞bn+1
B. an+1≥bn+1
C. an+1＜bn+1
D. an+1=bn+1
因为等差数列{an}和等比数列{bn}各项都是正数，且a1=b1，a2n+1=b2n+1，
所以an+1-bn+1=
a1+a2n+1
2-
b1•b2n+1=
a1+a2n+1−2
a1+a2n+1
2=
(
a1−
a2n+1)2
2≥0．
即 an+1≥bn+1．
故选 A．
所以an+1-bn+1=
a1+a2n+1
2-
b1•b2n+1=
a1+a2n+1−2
a1+a2n+1
2=
(
a1−
a2n+1)2
2≥0．
即 an+1≥bn+1．
故选 A．
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