证明:(1)∵an=a1*q^(n-1)
∴am=a1*q^(m-1)
ak=a1*q^(k-1)
at=a1*q^(t-1)
aman=a1*q^(n-1)*a1*q^(m-1)=a1^2*q^(m+n-2)
同理akat=a1^2*q^(k+t-2)
∵m+n=k+t,
∴a1^2*q^(m+n-2)=a1^2*q^(k+t-2)
∴aman=akat
(2)设四个正数分别为a1,a1q,a1q^2,a1q^3则百
a1*a1q*a1q^2*a1q^3=a1^4q^6=36
即a^2q^3=6
(1)
又a1q+a1q^2=5
即a1q(1+q)=5
(2)
(1)/(2)得a1=6(1+q)/(5q^2)代入(2)中百
6q^2-13q+6=0
解得q=2/3或q=3/2
当q=2/3时a1=3/2此时a1q=1
a1q^2=2/3
a1q^3=4/9
当q=3/2时a1=4/3此时a1q=2
a1q^2=3
a1q^3=9/2
故这四个数及公比分别为3/2,1,2/3,4/9和2/3或4/3,2,3,9/2和3/2
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