一道数列题，详细解答，急

如果q=1
5An=3 An=0.6
5(An)^2=1.8≠12
不符合题意。
∴q≠1

A1+A2+A3+A4+A5=A1×(1-q^5)/(1-q)=3
(A1)^2+(A2)^2+(A3)^3+(A4)^2+(A5)^2
=(A1)^2×(1-(q^2)^5)/(1-(q)^2)
=(A1)^2×(1-q^5)×(1+q^5)/[(1+q)(1-q)]
=[A1×(1-q^5)/(1-q)]×[A1×(1+q^5)/(1+q)]
=3[A1×(1+q^5)/(1+q)]=12
A1×(1+q^5)/(1+q)=4

A1-A2+A3-A4+A5
=A1×(1-(-q)^5)/(1-(-q))
=A1×(1+q^5)/(1+q)
=4

a1+a2+a3+a4+a5=3,...①
a1^2+a2^2+a3^2+a4^2+a5^2=12,...②
则a1-a2+a3-a4+a5=x...③
设①式等比为q,则②式为q^2，③式为-q
由等比求和公式，
①式可化为a1(1-q^5)/(1-q)=3
②式可化为a1^2(1-q^10)/(1-q^2)=12
③式可化为a1(1+q^5)/(1+q)=x
①×③刚好可以得到②式
即(a1(1-q^5)/(1-q))×(a1(1+q^5)/(1+q))=a1^2(1-q^10)/(1-q^2)=3x=12→x=4即③式为4