Question

设{an}是有正数组成的等比数列, Sn为其前n项和。已知a2a4=1,S3=7,则S5=

Answer 1

解：因为a2*a4=1,{an}是由正数组成的等比数列
所以a3=√(a2*a4)=1
又S3=7
故a1+a2+a3=a2/q+a2+a2*q=1/q+1+q=7
所以q^2-6q+1=0
故q=3+2√2或q=3-2√2

当q=3+2√2时
a1=a2/q=1/(3+2√2)=3-2√2
所以S5=a1*(1-q^5)/(1-q)=(3-2√2)*[1-(3+2√2)^5]/[1-(3+2√2)]
=123+82√2

当q=3-2√2时
a1=a2/q=1/(3-2√2)=3+2√2
所以S5=a1*(1-q^5)/(1-q)=(3+2√2)*[1-(3-2√2)^5]/[1-(3-2√2)]
=123-82√2

Answer 2

a2a4=1
a1qa1q^3=1
a1^2q^4=1
{an}是由正整数组成的等比数列
a1>0
q>0
a1q^2=1
S3=[a1(1-q)^3]/(1-q)=7
a1(1+q^2+q)=7
1+q^2+q=7/a1=7q^2
6q^2-q-1=0
(2q-1)(3q+1)=0
q=1/2
a1=4
S5={4(1-1/2)^5}/(1-1/2)=31/4

Answer 3

31/4