Question

在等比数列｛an}中，若a1+a2+a3+a4+a5=3，a6+a7+a8+a9+a10=9，则a11+a12+a13+a14+a15=?

Answer 1

3=a1+a2+a3+a4+a5=a1(1+q+q�0�5+q^3+q^4)……(1)
9=a6+a7+a8+a9+a10=a1q^5(1+q+q�0�5+q^3+q^4)……(2)
(2)÷(1)得：q^5=3
那么：a11+a12+a13+a14+a15
=(q^5)�0�5xa1(1+q+q�0�5+q^3+q^4)
=3�0�5x3
=27

Answer 2

a1(1+q+q^2+q^3+q^4)=3
a1*q^5(1+q+q^2+q^3+q^4)=9
所以q^5=3
a11+a12+a13+a14+a15=

a1*q^10(1+q+q^2+q^3+q^4)
=3*q^10
=3*(q^5)^2
=3*9
=27

Answer 3

a1+a2+a3+a4+a5=a1+a1q+a1q^2+a1q^3+a1q^4=a1(1+q+q^2+q^3+q^4)=3
a6+a7+a8+a9+a10=q^5(a1(1+q+q^2+q^3+q^4))=3q^5=9
q^5=3
a11+a12+a13+a14+a15=q^5(q^5(a1(1+q+q^2+q^3+q^4))=3*9=27