设an是公比为q的等比数列,4a1,3a2

S4=a1+a2+a3+a4=a2/q+a2+a2*q+a2*q^2S4/a2=1/q+1+q+q^2=7.5

第二题：1/(X-1)=1X>=2所以不等式解集为X=2第一题公比q若为正数的话,哪么应该大于1,因为要是q

S4=a1(1-q^4)/(1-q)=5a1(1-q^2)/(1-q)1+q^2=5q^2=4因为q

a1+a1q>2a1q^22q^2-q-1

充分不必要

首先得求的a1a4=5s2...a1q^3=5(a1+a1q)又.a3=a1q^2=2...所以.2q=5(a1+a1q)得.a1=(2q)/(5(1+q))又因为.a3=a1q^2=2得.q=1.2

等比数列an=a1*q^(n-1),Sn=a1(1-q^n)/(1-q)∴a3=2=a1*q^(3-1)=a1*q^2S4=5S2=>a1(1-q^4)/(1-q)=5*a1(1-q^2)/(1-q)

（Ⅰ）∵设{an}是公比为正数的等比数列∴设其公比为q，q＞0∵a3=a2+4，a1=2∴2×q2=2×q+4解得q=2或q=-1∵q＞0∴q=2∴{an}的通项公式为an=2×2n-1=2n（Ⅱ）∵

S4=a1(1-q4)/(1-q),S2=a1(1-q2)/(1-q),已知S4=5S2,则a1(1-q4)/(1-q)=5a1(1-q2)/(1-q),即q=±2,又公比q

(1)S1→3=a1(1+q+q^2)=a1*(1-q^3)/(1-q)S4→6=a4(1+q+q^2)=a1*(1-q^3)/(1-q)*q^3S7→9=a7(1+q+q^2)=a1*(1-q^3)

a2004和a2005是方程4x平方+8x+3=0的两根4x^2+8x+3=0(2x+3)(2x+1)=0x=-3/2x=-1/2∵q>1∴a2004=-1/2a2005=-3/2q=a2005/a2

4x²-8x+3=0（2x-3)(2x-1)=0x1=1/2,x2=3/2于是a2004=1/2,a2005=3/2,q=3a2006=a2004*q²,a2007=a2005*q

因为|q|>1所以|an|

S4=a1(1-q4)/(1-q),S2=a1(1-q2)/(1-q),已知S4=5S2,则a1(1-q4)/(1-q)=5a1(1-q2)/(1-q),即q=±2,又公比q

2a4=-a5+a62a4=-a4q+a4q^22a4=-a4q+a4q^2a4q^2-a4q-2a4=0a4(q^2-q-2)=0a4(q-2)(q+1)=0(q-2)(q+1)=0q=2或q=-1

Tn=1/a1+1/a2+……+1/anTn/q=1/a2+……+1/an+1/(q*an)Tn-Tn/q=1/a1-1/(q*an)Tn=q/a1(q-1)-1/an(q-1)

Sn=a1*(1-(根号2)^n)/(1-根号2)Tn=（17Sn-S2n）/an+1将Sn=a1*(1-(根号2)^n)/(1-根号2)an+1=a1*根号2^n带入其中求解,得(17-17根号2^

1.A1q^3+A1q^6=2A1q^9.解之得q^3=12.当q=1时A2=A1A5=A1A8=A1所以A2+A5=2A8所以a2,a8,a5成等差数列

设首项为a1，则s1=a1，s2=a1+a1qs3=a1+a1q+a1q2由于{Sn}是等差数列，故2（a1+a1q）=a1+a1+a1q+a1q2q2-q=0解得q=1．故答案为：1．

(1)令S=a1+a2+.+an,即S=a1+a1*q+.+a1*q^(n-1)则qS=a1*q+a1*q^2+a1*q^n故(1-q)S=a1-a1*q^n得S=a1(1-q^n)/(1-q)（2）