an = a1 + (n-1)*d a = a1 + 2nd bn = b1 * q^(n-1) b = b1 * q^(2n) a1 + 2nd = a1 * q^(2n) 2(a1 + nd) = a1 * [1 + q^(2n)] a = a1 + nd = a1 * [1 + q^(2n)]/2 b = a1 * q^n a - b = (a1 /2) * [q^(2n) - 2 q^n + 1] = (a1 /2) * (q^n -1)^2 所以a 大于
设{an}等差数列,{bn}是正项等比数列,且a1=b1,a(2n+1)=b（2n+1）,则a(n+1)与b（n+1）的
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