等比数列中,A1=-3 2,A7=-96

an=a1q^(n-1)a1+a3=30a1(1+q^2)=30(1)a5+a7=120a1q^4(1+q^2)=120(2)(2)/(1)q^4=40a9+a11=a1q^8(1+q^2)=30q^

由于是等比数列原式=(a4)^2+2a4*a6+(a6)^2=(a4+a6)^2=100

a5/a1=q^4a7/a3=q^4所以(a5+a7)/(a1+a3)=5/10=q^4=1/2a9/a5=q^4a11/a7=q^4所以（a9+a11)/(a5a+7)=q^4=1/2所以a9+a1

因为a3+a5+a7=9所以3*a5=9所以a5=3所以a5-a1=4d=3-1=2（d是公差）所以d=0.5所以an=a1+(n-1)d=1+(n-1)*0.5即an=0.5n+0.5字数不够bn过

a4=a1q^3a5=a1q^4a6=a1q^5a7=a1q^6a1q^3*a1q^6=32即a1^2q^9=32(1)a1q^4+a1q^5=12a1q^4(1+q)=12平方a1^2*q^8(1+

在等比数列｛an｝中,若a4*a7=32那么a5*a6=a4*a7=32又a5+a6=12解方程组得a5=4,a6=8或a5=8,a6=4①若a5=4,a6=8那么q=a6/a5=8/4=2所以a1=

==等比数列的性质没记住啊a1a7=a4²a2a6=a4²a3a5=a4²所以a1*a2*a3……a7=a4^7=3^7=2187

(1)a3=a1+2d,a7=a1+6d,所以a1*a7=a3*a3,即a1*（a1+6d）=（a1+2d）*（a1+2d）解得d=1(2)Sn=(1/2)n^2+(3/2)n,又a3=a1+2d=4

因为a1+a2+a3=7,a1a2a3=8又因为等比数列{an},那么a2*a2=a1a3,那么a1a2a3=a2a2a2=8,所以a2=2,那么a1+a3=5,同时a1a3=4所以a1=1,a3=4

a1=3/8a2=-1/8a3=1/24a4=-1/27a5=1/216a6=-1/648a7=1/1944a8=-1/5832a9=1/17496a10=-1/52488等比数列的比值为-1/3

240

a1+a3=a1+a1+2d=2a1+2d=10a5+a7=a1+4d+(a1+6d)=2a1+10d=5所以两式相减得10d-2d=8d=-5d=-5/8,所以a1=45/8所以a9+a11=a1+

A4=a1乘3个P,a6同理,二式除一式得P的三次方为2,又三式为二式乘P的3次方,所以为20

（1）a1(1+q^6)=65a1^2*q^6=64由a(n+1)

解析：∵a1=4,a7=4+6d,a10=4+9d∴a7^2=a1*a10,即（4+6d)^2=4(4+9d)∵d≠0∴d=-1/3即a1=4,a7=2,a10=1∴q=a2/a1=1/2∴Sn=4*

a4=a3.a5a1r^3=(a1r^2)(a1r^4)r^3=8r^6r^3(8r^3-1)=0r=(1/2)a7=a1r^6=8(1/2)^6=1/8

等比a3=a1q²a5=a3q²a7=a5q²所以(a3-a5+a7)/(a1-a3+a5)=q²=5/2同理aa9=a7q²所以(a5-a7+a9)

题目：等比数列an中,a1,1/2a3,2a2等差数列,（a8+a9）/（a6+a7）=?a1、a3/2、2a2成等差数列,则a3=a1+2a2a1q²=a1+2a1q（a1不等于0）整理,

∵a3-a5+a7=(a1-a3+a5)q^2∴q^2=(a3-a5+a7)/(a1-a3+a5)=5/2∴a5-a7+a9=(a3-a5+a7)q^2=5×5/2=25/2