请同学们思考以下问题:问题1:设数列{a_n}是正数等差数列,数列{b_n}是正数等比数列,且a_1=b_1,a(2n+1)=b_(2n+1).试比较a_(n+1)与b_(n+1)的大小关系.学生S_1很快给出了如下解法:因为a_(n+1)>0,b_(n+1)>0,所以,a_(n+1)=(a_1+a(2n+1))/2≥(a_1a_(2n+1))~(1/2)= Ask the students to think about the following questions: Question 1: Set the sequence {a_n} to be a positive array of equal numbers, and the sequence {b_n} to be a series of positive numbers, and a_1=b_1,a(2n+1)=b_(2n+1). Try to compare the size relationship between a_(n+1) and b_(n+1). Student S_1 quickly gave the following solution: because a_(n+1)> 0, b_(n+1)> 0, so A_(n+1)=(a_1+a(2n+1))/2≥(a_1a_(2n+1))~(1/2)=