等比数列方程组怎么解?

Question

有人会解这个方程组吗?求求大家了

Answer 1

(23)
(1)
an >0
an=a1.q^(n-1)
a1+a3=10
a1.(1+q^2) =10 (1)
a2+a3=6
a1.q(1+q) =6 (2)
(1)/(2)
(1+q^2)/[q(1+q)] = 5/3
3(1+q^2) = 5q(1+q)
2q^2+5q-3=0
(2q-1)(q+3)=0
q=1/2 or -3(rej)
from (1)
a1.(1+q^2) =10
a1.(1+1/4) =10
a1=8
an = 8.(1/2)^(n-1)
(2)
bn
= log<2>an
= log<2> [ 8.(1/2)^(n-1) ]
= log<2> 2^(4-n)
=4-n
=> { bn} 是等差数列
Sn
=b1+b2+...+bn
=n( 3+4-n)/2
=n(7-n)/2

Answer 2

一步一步减下去就可以了