资源描述:
《高考数学第一轮复习14等差、等比数列(二)》由会员上传分享,免费在线阅读,更多相关内容在学术论文-天天文库。
1、14.等差、等比数列(二)班级姓名一.选择题1.已知等差数列{an}满足,且an<0,则其前10项之和为( )(A)-9 (B)-11 (C)-13 (D)-152.在等差数列{an}中,若S9=18,Sn=240,an-4=30,则n的值为 ( )(A)14 (B)15 (C)16 (D)173.已知数列{an}是等比数列,Sn=48,S2n=60,则S3n为 ( )(A)75 (B)2880 (C) (D)634.若方程x2-5x+m=0与x2-10x+n=0的四个实根适当排
2、列后,恰好组成一个首项为1的等比数列,则m:n的值为 ( )(A)4 (B)2 (C) (D)5.等差数列{an}的公差d0,且a1,a3,a9成等比数列,则的值为( )(A) (B) (C) (D)二.填空题:6.设等差数列{an},{bn}的前n项和分别为An,Bn,且满足,则.7.一个等差数列的前12项和为354,前12项中,偶数项和与奇数项和之比为32:27,则公差d=.8.若等比数列{an}满足an>0,(nN﹡),公比q=2,且a1a2…a30=230,则a1a4a7…a
3、3k+1…a28的值是.9.在等差数列{an}中,若a10=0,则有等式a1+a2+…+an=a1+a2+…+a19-n(n<19,nN﹡)成立,类比上述性质,相应地,在等比数列{bn}中,若b9=1,则有等式成立.三.解答题:10.已知数列{an}中,a1=5,且an=3an-1+3n-1,(n=2,3,…)(1)试求a2,a3的值;(2)若存在实数,使得为等差数列,试求的值.11.设{an}是等差数列,,已知b1+b2+b3=,,求通项公式an.12.设数列{an}的首项a1=1,前n项和Sn满足关系式3tSn-(2t+3)Sn-1=3t(t>0,n=2,3,4,…)(
4、1)求证:数列{an}是等比数列;(2)设数列{an}的公比为f(t),作数列{bn},使b1=1,bn=,求数列{bn}的通项bn;(3)求和:b1b2-b2b3+b3b4-…+b2n-1b2n-b2nb2n+1.内部资料仅供参考9JWKffwvG#tYM*Jg&6a*CZ7H$dq8KqqfHVZFedswSyXTy#&QA9wkxFyeQ^!djs#XuyUP2kNXpRWXmA&UE9aQ@Gn8xp$R#͑Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89Am
5、YWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5ux^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wE
6、wZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmUE9aQ@Gn8xp$R#͑Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpaz
7、adNu##KN&MuWFA5ux^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z89AmYWpazadNu##KN&MuWFA5uxY7JnD6YWRrWwc^vR9CpbK!zn%Mz849Gx^Gjqv^$UE9wEwZ#Qc@UE%&qYp@Eh5pDx2zVkum&gTXRm6X4NGpP$vSTT#&ksv*3tnGK8!z8vG#tYM*Jg&6a*CZ7H$dq8KqqfHVZFedswSyXTy#&QA9w