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{a_n}是各项互不相等的正数等差数列，{b_n}是各项互不相等的正数等比数列，a₁=b₁，a_2n+1=b_2n+1，则（　　）
A. a_n+1＞b_n+1
B. a_n+1≥b_n+1
C. a_n+1＜b_n+1
D. a_n+1=b_n+1

因为等差数列{a_n}和等比数列{b_n}各项都是正数，且a₁=b₁，a_2n+1=b_2n+1，
所以a_n+1-b_n+1=
a1+a2n+1
2-
b1•b2n+1=
a1+a2n+1−2
a1+a2n+1
2=
(
a1−
a2n+1)2
2≥0．
即 a_n+1≥b_n+1．
故选 A．

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