Question

等差数列{an}中,前n项和为Sn且Sm/Sn=(m^2-2m)/(n^2-2n)则am/an=

Answer 1

解：∵{an}为等差数列
∴S（2m-1)/S(2n-1)
=｛[a1+a(2m-1)]（2m-1)/2｝/｛[a1+a(2n-1)]（2n-1)/2｝
=[am（2m-1)]/[an（2n-1)]
=[（2m-1)^2-2（2m-1)]/[(2n-1)^2-2(2n-1)]
=[（2m-1)（2m-3)]/[（2n-1)（2n-3)]
∴am/an=（2m-3)/（2n-3)

Answer 2

∵{an}为等差数列
∴Sm=ma1 + m(m-1)d/2 =[2ma1 + md(m-1)]/2
Sn=na1 + n(n-1)d/2 =[2na1 + nd(n-1)]/2
∵Sm/Sn=(m^2-2m)/(n^2-2n)
∴[2ma1 + md(m-1)] / [2na1 + nd(n-1)] = (m^2-2m) / (n^2-2n)
化简，整理后：(m-n)(2a1+d)=0
∵m≠n
∴d=-2a1
am=a1+(m-1)d=3a1-2a1m=a1(3-2m)
an=a1+(n-1)d=3a1-2a1n=a1(3-2n)
∴am/an=(3-2m)/(3-2n)

Answer 3

am/an=（2m-3)/（2n-3)