已知{an}是等比数列,an>0,a1a5+2a4^2=a3a7=36,则a3+a5=
问题描述:
已知{an}是等比数列,an>0,a1a5+2a4^2=a3a7=36,则a3+a5=
答
设初项为A1,公比Qan>0,a1a5+2a4^2+a3a7=36则有 A1*A1*Q^4+2(A1*Q^3)^2+A1*Q^2*A1*Q^6=36A1^2*(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)=6A1*Q^4+A1*Q^2=A5+A3=6所以a3+a5=6