Question

两正数数列{an} {bn}满足:an,bn,a(n+1)成等差数列,bn,a(n+1),b(n+1)成等比数列 a1=

Answer 1

①必要性：

设{an}成等差数列,公差为d,∵{an}成等差数列.

bn＝（a1+2a2+…+nan）/（1+2+3+…n）=[a1(1+2+...+n)+d(1*2+2*3+...+(n-1)*n)/(1+n...+n)=a1+2(n-1)d/3,

从而bn+1－bn=a1+2nd/3－a1－2(n－1)d/3=2d/3为常数.?

故{bn}是等差数列,公差为 2d/3.

②充分性:

设{bn}是等差数列,公差为d′,则bn=(n－1)d′?

∵bn*(1+2+…+n)=a1+2a2+…+nan ①

bn－1*(1+2+…+n－1)=a1+2a2+…+(n－1)an ②

①－②得：nan=n(n+1)bn/2-n(n-1)(bn－1)/2?

∴an=(n+1)bn/2-(n-1)(bn-1)/2=(n+1)[b1+(n-1)d']/2-(n-1)[b1+(n-2)d']/2=b1+3(n-1)d'/2 ,

从而得an+1－an= 3d′/2为常数,故{an}是等差数列.

综上所述,数列{an}成等差数列的充要条件是数列{bn}也是等差数列.