专题突破练12 等差、等比数列的综合问题1.(2022北京东城一模,文15)已知Sn是等差数列{an}的前n项和,且a3=-6,S5=S6.(1)求{an}的通项公式;(2)若等比数列{bn}满足b1=a2,b2=S3,求{bn}的前n项和.2.已知{an}是公差为3的等差数列,数列{bn}满足b1=1,b2=,anbn+1+bn+1=nbn.(1)求{an}的通项公式;(2)求{bn}的前n项和.3.(2022北京西城一模,文15)设等差数列{an}的公差不为0,a2=1,且a2,a3,a6成等比数列.(1)求{an}的通项公式;7\n(2)设数列{an}的前n项和为Sn,求使Sn>35成立的n的最小值.4.已知等比数列{an}的前n项和为Sn,a1=3,且3S1,2S2,S3成等差数列.(1)求数列{an}的通项公式;(2)设bn=log3an,求Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1.5.(2022北京顺义一模,文16)已知{an}是等差数列,{bn}是单调递增的等比数列,且a2=b2=3,b1+b3=10,b1b3=a5.7\n(1)求{an}的通项公式;(2)设cn=求数列{cn}的前n项和.6.设{an}是公比大于1的等比数列,Sn为数列{an}的前n项和,已知S3=7,且a1+3,3a2,a3+4构成等差数列.(1)求数列{an}的通项;(2)令bn=ln,n=1,2,…,求数列{bn}的前n项和Tn.7.(2022山西吕梁一模,文17)已知{an}是首项为1的等比数列,数列{bn}满足b1=2,b2=5,且anbn+1=anbn+an+1.(1)求数列{an}的通项公式;(2)求数列{bn}的前n项和.7\n8.(2022天津卷,文18)设{an}是等差数列,其前n项和为Sn(n∈N*);{bn}是等比数列,公比大于0,其前n项和为Tn(n∈N*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;(2)若Sn+(T1+T2+…+Tn)=an+4bn,求正整数n的值.参考答案专题突破练12 等差、等比数列的综合问题1.解(1)设等差数列{an}的公差为d.因为S5=S6,所以a6=a3+3d=0.因为a3=-6,所以d=2,a1=-10.所以an=2n-12.7\n(2)设等比数列{bn}的公比为q.由(1)可知,b1=-8,b2=-24,所以q=3.数列{bn}的前n项和为=4(1-3n).2.解(1)由已知,a1b2+b2=b1,b1=1,b2=,得a1=2.所以数列{an}是首项为2,公差为3的等差数列,通项公式为an=3n-1.(2)由(1)和anbn+1+bn+1=nbn得bn+1=,因此{bn}是首项为1,公比为的等比数列.记{bn}的前n项和为Sn,则Sn=.3.解(1)设等差数列{an}的公差为d,d≠0.∵a2,a3,a6成等比数列,∴=a2·a6,即(1+d)2=1+4d,解得d=2或d=0(舍去d=0),∴an=a2+(n-2)d=2n-3.(2)∵an=2n-3,∴Sn==n2-2n.依题意有n2-2n>35,解得n>7.故使Sn>35成立的n的最小值为8.4.解(1)∵3S1,2S2,S3成等差数列,∴4S2=3S1+S3,∴4(a1+a2)=3a1+(a1+a2+a3),即a3=3a2,∴公比q=3,∴an=a1qn-1=3n.(2)由(1)知,bn=log3an=log33n=n,∵b2n-1b2n-b2nb2n+1=(2n-1)·2n-2n(2n+1)=-4n,∴Tn=(b1b2-b2b3)+(b3b4-b4b5)+…+(b2n-1b2n-b2nb2n+1)=-4(1+2+…+n)=-4×=-2n2-2n.5.解(1)设等差数列{an}的公差为d,等比数列{bn}的公比为q.7\n由解得由解得所以an=2n-1.(2)设数列{cn}的前n项和为Sn,由(1)可知an=2n-1,bn=b1qn-1=3n-1.当n≤5时,Sn=a1+a2+…+an==n2.当n>5时,Sn=a1+a2+…+a5+b6+b7+…+bn==25+.6.解(1)由已知得解得a2=2.设数列{an}的公比为q,由a2=2可得a1=,a3=2q.又S3=7,所以+2+2q=7,即2q2-5q+2=0.解得q=2或q=.∵q>1,∴q=2,∴a1=1.故数列{an}的通项为an=2n-1.(2)由(1)得=23n,7\n∴bn=ln23n=3nln2.∵bn+1-bn=3ln2,∴数列{bn}为等差数列.∴Tn=b1+b2+…+bn=ln2.故Tn=ln2.7.解(1)把n=1代入已知等式得a1b2=a1b1+a2,∴a2=a1b2-a1b1=3a1.∴{an}是首项为1,公比为3的等比数列,即an=3n-1.(2)由已知得bn+1-bn==3,∴{bn}是首项为2,公差为3的等差数列,其通项公式为bn=3n-1,∴Sn=.8.解(1)设等比数列{bn}的公比为q.由b1=1,b3=b2+2,可得q2-q-2=0.因为q>0,可得q=2,故bn=2n-1.所以,Tn==2n-1.设等差数列{an}的公差为d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,从而a1=1,d=1,故an=n.所以,Sn=.(2)由(1),有T1+T2+…+Tn=(21+22+…+2n)-n=-n=2n+1-n-2.由Sn+(T1+T2+…+Tn)=an+4bn可得,+2n+1-n-2=n+2n+1,整理得n2-3n-4=0,解得n=-1(舍),或n=4.所以,n的值为4.7
