Question

等比数列{an}中的各项均为正数，且a5a6+a4a7=18，则log3a1+log3a2+...+log3a10=

Answer 1

解:
由于{an}为等比数列
则:a5a6=a4a7=a3a8=a2a9=a1a10

又a5a6+a4a7=18
则:
2a5a6=18
a5a6=9

则:
log3(a1)+log3(a2)+...+log3(a9)+log3(a10)
=log3[a1*a2*a3*...*a10]
=log3[(a1a10)*(a2a9)*...*(a5a6)]
=log3[9*9*...*9]
=log3[9^5]
=log3[3^10]
=10log3[3]
=10

祝你学习愉快

Answer 2

选B

a5a6+a4a7=a1^2*(q^9+q^9)=18, 故a1*q^4.5=3,log3a1+log3a2+...+log3a10=log3a1a2a3...a10=log3(a1^10*q^45)=log3(3^10)=10

满意望好和原创5快速采纳，多谢了~