s8=48 s12=168a1=

1.S8-S4=q^4S4即q^4=2S12-S8=q^8S4则S12=142.设an的公比为qS2n-Sn=q^nSnS3n-S2n=q^2nSn显然,Sn,S2n-Sn,S3n-S2n为公比为q^

1.用等比数列前n项和公式Sn=a1(1-q^n)/(1-q),(q≠1)带入S4/S8=1/4解得q^4=3或q^4=1（即q=-1）将q^4=3带入S12/S16=(1-q^12)/(1-q^16

q^7=a8/a1=81÷1/27=3^7q=3所以S8=a1*)1-3^8)/(1-3)=3280/27再问：你们答案都不一样到底哪个是对的啊？

等比!a1+a1q³=18a1q+a1q²=12相除(q+1)(q²-q+1)/q(q+1)=18/122q²-2q+2=3qq是整数所以q=2a1=18/(1

a1+a1q^3=18a1q+a1q^2=12因为q为整数,所以q=2,a1=2s8=2x(1-2^8)/(1-2)=510

S3=a1+a2+a3=a1+a1+d+a1+2d=3a1+3dS8=a1+a2+a3+...+a8=a1+a1+d+a1+2d+...+a1+7d=8a1+(d+2d+...+7d)=8a1+28d

解题思路：数列································解题过程：

设数列｛an｝公差为d.S4=a1+a2+a3+a4=6S8-S4=a5+a6+a7+a8S12-S8=a9+a10+a11+a12易知a1+a2+a3+a4,a5+a6+a7+a8,a9+a10+a

设公比为q.a1+a2=a1(1+q)=3(1)a3+a4=a3(1+q)=a1q^2(1+q)=6(2)(2)/(1)q^2=2a1(1+q)=3a1=3/(1+q)S8=a1(q^8-1)/(q-

s8=8a1+28d=48,2a1+7d=12(1)s12=12a1+66d=108,2a1+11d=18(2)（1）-（2）解得：4d=6,d=1.5把d=1.5代入（1）得a1=0.75没分就给个

∵等比数列{an}中，S4=3，S12-S8=12，∴a9+a10+a11+a12=q4（a5+a6+a7+a8）=q4（S8-S4）=6•q4=12，∴q4=2，∴a5+a6+a7+a8=q4（a1

S12=a1+a12=(a1+a12)*6=[a1+a1+(12-1)*d]*6=(2a1+11d)*68S4=8*[(a1+a4)*2]=8*{[a1+a1+(4-1)*d]*2}=8*[(2a1+

等差数列{an}s4,s8-s4,s12-s8也成等差数列2(s8-s4)=s4+s12-s82s8-2s4=s4+s12-s8s12=3s8-3s4s12=3(s8-s4)s12=3*(4-8)s1

等差数列S4,S8-S4,S12-S8也为等差,根据等差中项等于两边项之和的二倍得S12=12

S8=48,S12=168S12-S8=120即A9+A10+A11+A12=120所以有：A4+A5=48/4=12A10+A11=120/2=60所以公差d=(60-12)/(6+6)=4所以A4

因为数列{an}是等比数列所以S4,S8-S4,S12-S8也成等比数列所以(6-2)^2=2*(S12-6)所以S12=14故选D

因为S4,S8-S4,S12-S8也成等比数列所以（S8-S4）²=S4×（S12-S8）即(6-2)^2=2*(S12-6)再问：（S8-S4）²=S4×（S12-S8）这是什么

S8=S12所以a9+a10+a11+a12=0a9+a12=a10+a11所以a10+a11=0所以a10=-a11又a10所以a10与a11一正一负（不可能是都为0）且a100所以当n=10时,S

S4=S12(a1+a4)*4/2=(a1+a12)*12/22a1+2a4=6a1+6a12a1+a1+3d=3a1+3a1+33d33d-3d=a1+a1-3a1-3a130d=-4a1=-60d

应该是等差数列吧.因为2）中有叫你求公差d1）a1+14d=33a1+44d=153得d=4,a1=-23所以a61=2172)8a1+36d=4812a1+78d=168a1=-12,d=43)a1