公差不为0的 等差数列an中a1.a3.a5成等比,求(a1+a3+a9)/(a2+a4+a10)如题.已知等比数列{an}的前N项和Sn=(1/2)^n+a.则a=？第一题答案是13/16诶=

题目有误，应该是a1,a3,a9成等比
a3=a1+2d
a9=a1+8d
(a3)^2=a1×a9
(a1+2d)^2=a1×(a1+8d)
(a1)^2+4a1d+4d^2=(a1)^2+8a1d
d^2=a1d
d不等于0
a1=d
(a1+a3+a9)/(a2+a4+a10)
=(d+3d+9d)/(2d+4d+10d)
=13/16
问题补充解答
an=Sn-S(n-1) (n>=2)
=(1/2)^n+a-[(1/2)^(n-1)+a]
=(1/2)^n-(1/2)^(n-1)
=-(1/2)^n
由此得出公比q=1/2
a2=-(1/2)^2=-1/4
a1=S1=1/2+a
(1/2+a)×(1/2)=-1/4
1/2+a=-1/2
a=-1

你的答案似乎不对,因为我做过这道题三遍.
1.a1,a3,a5成等比
则：a3^2=a1*a5
又a1,a3,a5是等差数列{an}中的项
则:a3=a1+2d
a5=a1+4d
则有:(a1+2d)^2=a1(a1+4d)
a1^2+2a1d+4d^2=a1^2+4a1d
4d^2-2a1d=0
2d^2-a1d=0
d(2d-a1)=0
又d不为0
则:2d-a1=0
a1=2d
则(a1+a3+a9)/(a2+a4+a10)
=[a1+a1+2d+a1+8d]/[a1+d+a1+3d+a1+9d]
=[3*2d+10d]/[3*2d+13d]
=16/19
2.an=Sn-S(n-1) (n>=2)
=(1/2)^n+a-[(1/2)^(n-1)+a]
=(1/2)^n-(1/2)^(n-1)
=(1/2)^n-[(1/2)^n]/[(1/2)]
=(1/2)^n-2*[(1/2)^n]
=(1-2)*(1/2)^n
=-(1/2)^n
又a1=S1=(1/2)^1+a
=1/2+a
则:an=-(1/2)^n (n>=2)
=1/2+a (n=1)
又{an}为等比数列
则:a1=-(1/2)^1=1/2+a
-1/2=1/2+a
则:a=-1