已知等比数列{an}满足:a1+a2+a3+...+a99=100 1/a1+1/a2+1/a3+...+1/a99=1/100,则a1a2a3...a99=

题目

已知等比数列{an}满足:a1+a2+a3+...+a99=100 1/a1+1/a2+1/a3+...+1/a99=1/100,则a1a2a3...a99=
如题

答案

A1+A2+……+A99=A1×(1-q^99)/(1-q)=100
数列{1/An}也是等比数列,首项为1/A1,公比为1/q
1/A1+1/A2+……+1/A99=1/A1×(1-1/q^99)/(1-1/q)=1/100
两式相除
[A1×(1-q^99)/(1-q)]/[1/A1×(1-1/q^99)/(1-1/q)]
=A1^2×[(1-q^99)/(1-q)]/[(q^99/q)×(q^99-1)/(q-1)]
=A1^2×q^98
=(A1×q^49)^2=100/(1/100)=10000
A1×q^49=±100
A1×A2×……×A99
=A1^99×q^(0+1+2+……+98)
=A1^99×q^(98×99/2)
=(A1×q^49)^99
=(±100)^99
=±10^198