已知{an}是等比数列,an>0,a1a5+2a4^2=a3a7=36,则a3+a5=

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$已知{an}是等比数列,an>0,a1a5+2a4^2=a3a7=36,则a3+a5=$

已知{an}是等比数列,an>0,a1a5+2a4^2=a3a7=36,则a3+a5=
已知{an}是等比数列,an>0,a1a5+2a4^2=a3a7=36,则a3+a5=

已知{an}是等比数列,an>0,a1a5+2a4^2=a3a7=36,则a3+a5=
设初项为A1,公比Q
an>0,a1a5+2a4^2+a3a7=36
则有 A1*A1*Q^4+2(A1*Q^3)^2+A1*Q^2*A1*Q^6=36
A1^2*(Q^4+2Q^6+Q^8)=36
(A1*Q^2*(Q^2+1))^2=36
因为AN>0
所以A1*Q^2*(Q^2+1)=6
A1*Q^4+A1*Q^2=A5+A3=6
所以a3+a5=6

设公比为q,由a1a5+2a4^2=a3a7=36得
a1^2*(q^4+2q^6)=a1^2*q^8=36,
∴1+2q^2=q^4,q^4-2q^2-1=0,
q^2=1+√2，
a1=6/q^4,
∴a3+a5=a1(q^2+q^4)=6(1/q^2+1)
=6√2.

{an}是等比数列，则a3 ^2=a1a5, a4 ^2=a3a5, a5 ^2=a3a7
an>0
所以a3+a5=√（a3+a5）^2=√（a3 ^2+2a3a5+a5^2)
=√（a1a5+2a4 ^2+a3a7)
=√36
=6

等比数列an>0,a1a5+2a4^2=a3a7=36
2a4^2=2a3*a5,a3a7=a5^2,a1a5=a3^2
a1a5+2a4^2=a3a7=a3^2+2a3a5=a5^2=36
a5=6,a3^2+2a3a5+a5^2=36+36
a3+a5=6根号2

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