第三节等比数列及其前n项和A组基础题组1.已知等比数列{an}的前n项和为Sn,且a1+a3=52,a2+a4=54,则Snan=()A.4n-1B.4n-1C.2n-1D.2n-1答案D设等比数列{an}的公比为q, {a1+a3=52,a2+a4=54,∴{a1+a1q2=52,①a1q+a1q3=54,②由①÷②可得1+q2q+q3=2,∴q=12,代入①解得a1=2,∴an=2×(12)n-1=42n,Sn=2×[1-(12)n]1-12=4(1-12n),∴Snan=4(1-12n)42n=2n-1,选D.2.(2019湖北武汉调研)设公比为q(q>0)的等比数列{an}的前n项和为Sn.若S2=3a2+2,S4=3a4+2,则a1=()A.-2B.-1C.12D.23答案B由S2=3a2+2,S4=3a4+2作差可得a3+a4=3a4-3a2,即2a4-a3-3a2=0,所以2q2-q-3=0,解得q=32或q=-1(舍).将q=32代入S2=3a2+2得,a1+32a1=3×32a1+2,解得a1=-1.3.(2018湖北武汉联考)已知{an}为等比数列,a4+a7=2,a5a6=-8,则a1+a10=()A.7B.5C.-5D.-7答案D由{a4+a7=2,a5a6=-8得{a4+a7=2,a4a7=-8,解得{a4=4,a7=-2或{a4=-2,a7=4.若a4=4,a7=-2,则a1=-8,a10=1,此时a1+a10=-7;若a4=-2,a7=4,则a1=1,a10=-8,此时a1+a10=-7.故选D.4.已知等比数列{an}的前n项和为Sn=a·2n-1+16,则a的值为()A.13B.12C.-13D.-12答案C由题意得a1=a+16,a2=S2-S1=a,a3=S3-S2=2a,又a22=a1a3,所以a2=(a+16)·2a,解得a=-13或a=0(舍去),故选C.5.已知递增的等比数列{an}的公比为q,其前n项和Sn<0,则()A.a1<0,01C.a1>0,00,q>1答案A Sn<0,∴a1<0,又数列{an}为递增等比数列,∴an+1>an,且|an|>|an+1|,则-an>-an+1>0,则q=-an+1-an∈(0,1),∴a1<0,01,令bn=an+1(n=1,2,…),若数列{bn}有连续四项在集合{-53,-23,19,37,82}中,则q等于()A.-12B.12C.-32D.32答案C{bn}有连续四项在{-53,-23,19,37,82}中且bn=an+1,则{an}有连续四项在{-54,-24,18,36,81}中. {an}是等比数列,等比数列中有负数项,则q<0,且负数项为相隔两项,∴等比数列各项的绝对值递增或递减.按绝对值的顺序排列上述数值:18,-24,36,-54,81,相邻两项相除:-2418=-43,36-24=-32,-5436=-32,81-54=-32,则可得-24,36,-54,81是{an}中连续的四项.q=-32或q=-23( |q|>1,∴此种情况舍去),∴q=-32.故选C.2.在递增的等比数列{an}中,已知a1+an=34,a3·an-2=64,且前n项和为Sn=42,则n=.答案3解析设递增等比数列{an}的公比为q(q>0),a3an-2=a1an=64,a1+an=34,解得a1=2,an=32,则Sn=a1-anq1-q=42,解得q=4,则qn-1=4n-1=ana1=16,解得n=3.3.(2018四川...
