Question

已知等差数列{an}和等比数列{bn}各项都是正数，且a1=b1，a2n+1=b2n+1，那么，一定有（　　）A．an+1≤bn+

已知等差数列{an}和等比数列{bn}各项都是正数，且a1=b1，a2n+1=b2n+1，那么，一定有（　　）A．an+1≤bn+1B．an+1≥bn+1C．an+1＞bn+1D．an+1＜bn+1

Answer 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 B．