设{an}为等差数列,{bn}为等比数列又a1=b1,a3=b3,a7=b5.比较的a15和b7大小关系,给出证明

{an}为等差数列,公差为d
a3=a1+2d a7=a1+6d a15=a1+14d
{bn}为等比数列,公比为q
b3=b1q^2 b5=b1q^4 b7=b1q^6
∵a1=b1 a3=b3
∴a1+2d=b1q^2
b1+2d=b1q^2
2d=b1(q^2-1)——(1)
而a7=b5
则a1+6d=b1q^4
b1+6d=b1q^4
6d=b1(q^4-1)——(2)
(2)/(1)得：
3=(q^4-1)/(q^2-1)
q^2+1=3
q^2=2
2d=b1(q^2-1)=(2-1)b1=b1
∴a15=a1+14d=b1+7(2d)=b1+7b1=8b1
b7=b1q^6=b1(q^2)^3=(2^3)b1=8b1
∴a15=b7