1.等比数列{an}的前n项和为Sn,且4a1,2a2,a3成等差数列.若a1=1,则S4等于( )
A.7 B.8 C.15 D.16
解析:设等比数列的公比为q,
则由4a1,2a2,a3成等差数列,得4a2=4a1+a3,
∴4a1q=4a1+a1q2.∴q2-4q+4=0.
∴q=2.∴S4= =15.
答案:C
2.设Sn为等比数列{an}的前n项和,已知3S3=a4-2,3S2=a3-2,则公比q等于( )
A.3 B.4 C.5 D.6
解析:由题意,得3S3-3S2=(a4-2)-(a3-2),
则3a3=a4-a3,即a4=4a3,故q= =4.
答案:B
3.设Sn为等比数列{an}的前n项和,8a2+a5=0,则 等于( )
A.11 B.5 C.-8 D.-11
解析:设{an}的公比为q.
∵{an}为等比数列,且8a2+a5=0,
∴8a2+a2q3=0∵a2≠0,∴q3=-8.∴q=-2,
==-11.
