=3,而==a5=a1·q4=1×32=9.3.(2017·全国Ⅲ)等差数列{an}的首项为1,公差不为0.若a2,a3,a6成等比数列,则{an}的前6项和为( )A.-24B.-3C.3D.8答案 A解析由已知条件可得a1=1,d≠0,由a=a2a6,可得(1+2d)2=(1+d)(1+5d),解得d=-2或d=0(舍).所以S6=6×1+=-24.4.一个等比数列的前三项的积为2,最后三项的积为4,且所有项的积为64,则该数列的项数是( )A.13?B.12C.11?D.10答案 B解析设等比数列为{an},其前n项积为Tn,由已知得a1a2a3=2,anan-1an-2=4,可得(a1an)3=2×4,a1an=2,∵Tn=a1a2…an,∴T=(a1a2…an)2=(a1an)(a2an-1)…(ana1)=(a1an)n=2n=642=212,∴n=12.5.(2018·荆州质检)已知数列{an}满足=25·5an,且a2+a4+a6=9,则(a5+a7+a9)等于( )A.-3B.3C.-D.答案 A解析∵=25·=,∴an+1=an+2,∴数列{an}是等差数列,且公差为2.∵a2+a4+a6=9,∴3a4=9,a4=3.∴====-3.6.(2018·吉林调研)已知等差数列{an}的公差不为0,a1=1,且a2,a4,a8成等比数列,设{an}的前n项和为Sn,则Sn=________.答案(n∈N*)解析设等差数列{an}的公差为d.∵a2,a4,a8成等比数列,∴a=a2·a8,即(a1+3d)2=(a1+d)·(a1+7d),∴(1+3d)2=(1+d)·(1+7d),解得d=1或d=0(舍).∴Sn=na1+d=(n∈N*).7.(2018·资阳模拟)等差数列{an}的前n项和为Sn,若a2=8,且Sn≤S7,则公差d的取值范围是________.
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