要详解已知等比数列,（an）为递增数列,且a5方=A10,2（an+a(n+2）)=5a1,求通项公式

an=a1q^(n-1)
(a5)^2 = a10
(a1)^2.q^8 = a1q^9
a1= q
2[an+a(n+2)]=5a1
an+a(n+2)=5q/2
a(n+2) -5q/4= -(an -5q/4)
[a(n+2) -5q/4] /(an -5q/4) = -1
when n is odd
(an -5q/4)/(a1 -5q/4) = (-1)^[(n-1)/2]
an -5q/4 = -(q/4).(-1)^[(n-1)/2]
an = (5q/4) - (q/4).(-1)^[(n-1)/2]
n=3
a3= 5q/4 +q/4 = 3q/2 = q^3
q(q^2- 3/2) =0
q= √(3/2)
an = a1q^(n-1)
=(3/2)^(n/2)