怎么求等差数列的公比

Question

如题所述

Answer 1

方法1：
:∵{an}与{bn}是等差数列
∴Sn=[n(a1+an)]/2
Tn=[n(b1+bn)]/2
∴Sn/Tn=(a1+an)/(b1+bn)
∵等差数列{an}与{bn}的前n项和的比为2n:(3n+1)
∴(a1+an)/(b1+bn)=2n:(3n+1)
假设(n+1)/2 =k {(n+1)/2为项数}
则n=2k-1
则ak/bk = 2(2k-1)/[3(2k-1)+1]
=(2k-1)/(3k-1)
所以an/bn=(2n-1)/(3n-1)

方法2:
Sn/Tn=2n/(3n+1)
S(2n-1)/T(2n-1)=2(2n-1)/[3(2n-1)+1]=(2n-1)/(3n-1)
[a1+a（2n-1）]/[b1+b(2n-1)]=(2n-1)/(3n-1)
2an/2bn=(2n-1)/(3n-1),
an/bn=(2n-1)/(3n-1)