高考数学专题复习全集-21数列的应用.pptx

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,数列的应用,典型例题,解:,设第二个数为,a,则第三个数为,12,-,a,.,前三个数成等差数列,第一种数为 3a-12.,从而,第四个数为16,-,(3,a,-,12)=28,-,3,a,.,依题意得,:(12,-,a,),2,=,a,(,28,-,3,a,).,化简整顿得 a2-13a+36=0.,解得,a,=,4,或,9,.,这四个数分别为,0,4,8,16,或,15,9,3,1.,1.有四个数,前三个数成等差数列,后三个数成等比数列,并且第一种数与第四个数的和是 16,第二个数与第三个数的和是 12,求这四个数.,a,2,=1 从而,a,1,=1,-,d,a,3,=1+,d,.,整顿得 4(2d)2-17(2d)+4=0.,故,a,n,=2,n,-,3,或,a,n,=,-,2,n,+5.,2.,设,a,n,是等差数列,b,n,=(),a,n,已知,b,1,+,b,2,+,b,3,=,b,1,b,2,b,3,=,求等差数列,a,n,.,1,2,8,21,1,8,解:,设,a,n,的公差为,d,.,b,1,b,3,=(),a,1,(),a,3,=(),a,1,+,a,3,=(),2,a,2,=,b,2,2,1,2,1,2,1,2,1,2,由,b,1,b,2,b,3,=,得,b,2,3,=.,1,8,1,8,b,2,=,.,1,2,又由,b,1,+,b,2,+,b,3,=,得,(),1,-,d,+(),1+,d,=.,8,21,1,2,1,2,1,2,8,21,解得,2,d,=2,2,或,2,-,2,.,d,=2,或,-,2.,当,d,=2,时,a,n,=,a,2,+(,n,-,2),d,=1+2,n,-,4=2,n,-,3;,当,d,=,-,2,时,a,n,=,a,2,+(,n,-,2),d,=1,-,2,n,+4=,-,2,n,+5.,f,(,x,)=2,-,10,4,x,.,(2)由已知,a,n,=log,2,f,(,n,)=log,2,(2,-,10,4,n,)=2,n,-,10.,3.已知函数,f,(,x,)=,a,b,x,的图象过点,A(4,),和,B(5,1).(1)求函数,f,(,x,),的解析式;(2)记,a,n,=log,2,f,(,n,),n,为正整数,S,n,是数列,a,n,的前,n,项和,解关于,n,的不等式,a,n,S,n,0;(3)对于(2)中的,a,n,与,S,n,整数,10,4,是否为数列,a,n,S,n,中的项?若是,则求出相应的项数;若不是,则说明理由.,1,4,解:,(1)由已知,a,b,4,=,a,b,5,=1,1,4,解得,b,=4,a,=2,-,10,.,S,n,=,n,(,n,-,9).,a,n,S,n,=2,n,(,n,-,5)(,n,-,9).,n,N,*,由,a,n,S,n,0,得,(,n,-,5)(,n,-,9),0.,解得,5,n,9,n,N,*,.,n,=5,6,7,8,9.,(3),a,1,S,1,=64,a,2,S,2,=84,a,3,S,3,=72,a,4,S,4,=40;,当,5,n,9,时,a,n,S,n,0;,当,10,n,22,时,a,n,S,n,a,22,S,22,=972410,4,;,故整数,10,4,不是数列,a,n,S,n,中的项.,解:,(1)由已知数列,a,n,+1,-,a,n,是首项为,-,2,公差为,1,的等差数列.,a,n,+1,-,a,n,=(,a,2,-,a,1,)+(,n,-,1),1=,n,-,3.,a,n,=,a,1,+(,a,2,-,a,1,)+(,a,3,-,a,2,)+,+(,a,n,-,a,n,-,1,),4.设数列,a,n,和,b,n,满足,a,1,=,b,1,=6,a,2,=,b,2,=4,a,3,=,b,3,=3,且数列,a,n,+1,-,a,n,(,n,N,*,),是等差数列,数列,b,n,-,2,(,n,N,*,),是等比数列.(1)求数列,a,n,和,b,n,的通项公式;(2)是否存在,k,N,*,使,a,k,-,b,k,(0,)?若存在,求出,k,若不存在,说明理由,.,1,2,a,n,-,a,n,-,1,=,n,-,4(,n,2).,=6+(,-,2)+(,-,1)+0+1+2+,+(,n,-,4),=,(,n,2,-,7,n,+18)(,n,2).,1,2,而,a,1,=2,亦适合上式,=,(,n,2,-,7,n,+18)(,n,N,*,).,1,2,a,n,又数列,b,n,-,2,是首项为,b,1,-,2=4,公比为,的等比数列,1,2,b,n,-,2=4(,),n,-,1,=(,),n,-,3,.,1,2,1,2,b,n,=(,),n,-,3,+2.,1,2,故数列,a,n,和,b,n,的通项公式分别为:,a,n,=,(,n,2,-,7n+18),1,2,b,n,=(,),n,-,3,+2.,1,2,解:,(2)显然当,k,=1,2,3,时,a,k,-,b,k,=0,不适合题意;,数列,a,k,是递增数列,b,k,是递减数列.,不,存在,k,N,*,使,a,k,-,b,k,(0,),.,1,2,当,k,4,时,a,k,=,(,k,2,-,7,k,+18),b,k,=(,),k,-,3,+2,1,2,1,2,数列,a,k,-,b,k,是递增数列.,a,k,-,b,k,a,4,-,b,4,=3,-,(+2)=,1,2,1,2,(0,).,1,2,4.设数列,a,n,和,b,n,满足,a,1,=,b,1,=6,a,2,=,b,2,=4,a,3,=,b,3,=3,且数列,a,n,+1,-,a,n,(,n,N,*,),是等差数列,数列,b,n,-,2,(,n,N,*,),是等比数列.(1)求数列,a,n,和,b,n,的通项公式;(2)是否存在,k,N,*,使,a,k,-,b,k,(0,)?若存在,求出,k,若不存在,说明理由,.,1,2,5.,已知,等比数列,a,n,的各项均为正数,公比,q,1,数列,b,n,满足,b,1,=20,b,7,=5,且,(,b,n,+1,-,b,n,+2,)log,m,a,1,+(,b,n,+2,-,b,n,)log,m,a,3,+(,b,n,-,b,n,+1,)log,m,a,5,=0.(1)求数列,b,n,的通项公式;(2)设,S,n,=|,b,1,|+|,b,2,|+,+|,b,n,|,求,S,n,.,解:(1)将 logma3=logma1+2logmq,logma5=logma1+4logmq 代入已,知等式整顿得:,2(,b,n,-,2,b,n,+1,+,b,n,+2,)log,m,q,=0.,b,n,-,2,b,n,+1,+,b,n,+2,=0.,q,1,log,m,q,0.,即,b,n,+,b,n,+2,=2,b,n,+1,.,数列,b,n,是等差数列.,设其公差为,d,5,2,-,.,b,n,=20+(,n,-,1)(,-,).,5,2,即,b,n,=,-,n,+.,5,2,2,45,则由,b,7,=,b,1,+6,d,可得,d,=,解:,(2)令,b,n,=0,得,n,=9.,当,n,9,时,b,n,0.,则,S,n,=,b,1,+,b,2,+,+,b,n,=20,n,+,(,-,),n,(,n,-,1),2,5,2,=,-,n,2,+,n,.,5,4,4,85,当,n,9,时,b,n,9.,5,4,4,85,-,n,2,+,n,n,9,5,4,4,85,S,n,=,5.,已知,等比数列,a,n,的各项均为正数,公比,q,1,数列,b,n,满足,b,1,=20,b,7,=5,且,(,b,n,+1,-,b,n,+2,)log,m,a,1,+(,b,n,+2,-,b,n,)log,m,a,3,+(,b,n,-,b,n,+1,)log,m,a,5,=0.(1)求数列,b,n,的通项公式;(2)设,S,n,=|,b,1,|+|,b,2,|+,+|,b,n,|,求,S,n,.,6.设,a,0,为常数,且,a,n,=3,n,-,1,-,2,a,n,-,1,(,n,N,*,).(1)证明:对任意,n,1,a,n,=,3,n,+(,-,1),n,-,1,2,n,+(,-,1),n,2,n,a,0,;(2)假设对于任意,n,1,a,n,a,n,-,1,求,a,0,的取值范围.,1,5,(1),证:,由,a,n,=3,n,-,1,-,2,a,n,-,1,知:,3,+1,2,令,=,-得,=,-,.,1,5,则,a,n,-,3,n,是以,a,0,-,为首项,公比为,-,2,的等比数列.,1,5,1,5,a,n,-,3,n,=(,a,0,-,)(,-,2),n,.,1,5,1,5,即,a,n,=,3,n,+(,-,1),n,-,1,2,n,+(,-,1),n,2,n,a,0,.,1,5,(2),解:,由,a,n,a,n,-,1,及,a,n,=3,n,-,1,-,2,a,n,-,1,知:,a,n,-,a,n,-,1,=3,n,-,1,-,3,a,n,-,1,0.,a,n,-,1,0.,0,a,n,-,1,30,成立的,n,的最小值.,1,2,解:,(1)由已知可,设,等比数列,a,n,的公比为,q,依题意得:,a,1,q,+,a,1,q,2,+,a,1,q,3,=28,a,1,q,+,a,1,q,3,=2(,a,1,q,2,+2),解得或,(,舍去,),a,1,=2,q,=2,a,1,=32,q,=,1,2,a,n,=2,2,n,-,1,=2,n,.,即,a,n,的通项公式为,a,n,=2,n,.,(2),b,n,=,a,n,log,a,n,=,-,n,2,n,1,2,-,S,n,=1,2+2,2,2,+3,2,3,+,+,n,2,n,.,-,2,S,n,=1,2,2,+2,2,3,+3,2,4,+,+,n,2,n,+1,.,S,n,=2+2,2,+2,3,+,+2,n,-,n,2,n,+1,=2,n,+1,-,2,-,n,2,n,+1,.,为,使,S,n,+,n,2,n,+1,30,成立,应有,2,n,+1,32.,n,4.,使,S,n,+,n,2,n,+1,30,成立的,n,的最小值为,5.,S,n,=,-,(1,2+2,2,2,+3,2,3,+,+,n,2,n,).,9.以数列,a,n,的任意相邻两项为坐标的点,P,n,(,a,n,a,n,+1,)(n,N,*,)均在一次函数,y,=2,x,+,k,的图象上,数列,b,n,满足条件:,b,n,=,a,n,+1,-,a,n,(,n,N,*,b,1,0).,(1)求证:数列,b,n,是等比数列;,(2),设数列,a,n,b,n,的前,n,项和分别为,S,n,T,n,若,S,6,=,T,4,S,5,=,-,9,求,k,的值.,(1)证:,P,n,(,a,n,a,n,+1,),(,n,N,*,),均在一次函数,y,=2,x,+,k,的图象上,a,n,+1,=2,a,n,+,k,即:,a,n,+1,+,k,=2(,a,n,+,k,).,又,b,n,=,a,n,+1,-,a,n,=,a,n,+,k,则,b,n,+1,=,a,n,+1,+,k,b,n,+1,b,n,=,=2.,a,n,+1,+,k,a,n,+,k,数列,b,n,是等比数列.,解得,:,k,=8.,(2)解:,b,1,=,a,1,+,k,b,n,=(,a,1,+,k,),2,n,-,1,a,n,=,b,n,-,k,=(,a,1,+,k,),2,n,-,1,-,k,S,6,=,T,6,-,6,k,=(,a,1,+,k,)(2,6,-,1),-,6,k,=63,a,1,+5,k,T,4,=(,a,1,+,k,)(2,5,-,1)=15(,a,1,+,k,),S,5,=31,a,1,+26,k,=,-,9,S,6,=,T,4,a,1,=,-,k,7,8,10.(1)已知数列,c,n,其中,c,n,=2,n,+3,n,且数列,c,n,+1,-,pc,n,为等比数列,求常数,p,;,(2)设,a,n,b,n,是公比不相等的两个等比数列,c,n,=,a,n,+,b,n,证明:数列,c,n,不是等比数列.,(1),解:,数列,c,n,+1,-,pc,n,为等比数列,(,c,n,+1,-,pc,n,),2,=(,c,n,+2,-,pc,n,+1,)(,c,n,-,pc,n,-,1,).,又,c,n,=2,n,+3,n,2,n,+1,+3,n,+1,-,p,(2,n,+3,n,),2,=2,n,+2,+3,n,+2,-,p,(2,n,+1,+3,n,+1,)2,n,+3,n,-,p,(2,n,-,1,+3,n,-,1,).,即,(2,-,p,)2,n,+(3,-,p,)3,n,2,=(2,-,p,)2,n,+1,+(3,-,p,)3,n,+1,(2,-,p,)2,n,-,1,+(3,-,p,)3,n,-,1,.,整理得 (2,-,p,)(3,-,p,),2,n,3,n,0.,1,6,解得,p,=2,或,3.,(2),证:,设,a,n,b,n,的公比,分别为,p,q,p,q,.,为证,c,n,不是等比数列,只须证,c,2,2,c,1,c,3,.,事实上,c,2,2,=(,a,1,p,+,b,1,q,),2,=,a,1,2,p,2,+,b,1,2,q,2,+2,a,1,b,1,pq,c,1,c,3,=(,a,1,+,b,1,)(,a,1,p,2,+,b,1,q,2,)=,a,1,2,p,2,+,b,1,2,q,2,+,a,1,b,1,(,p,2,+,q,2,).,p,q,p,2,+,q,2,2,pq,.,又,a,1,b,1,不为零,c,2,2,c,1,c,3,.,故,c,n,不是等比数列.,11.设等比数列 an 的各项为实数,前 n 项的和为Sn,公比为q.(1)若 S5,S15,S10 成等差数列,求证:2S5,S10,S20-S10 成等比数列;(2)若 2S5,S10,S20-S10 成等比数列,试问若 S5,S15,S10一定成等差数列吗?请阐明理由.,(1),证:,由已知,q,1,(,若,q,=1,则,S,5,=5,a,1,S,15,=15,a,1,S,10,=10,a,1,不满足,S,5,S,15,S,10,成等差数列,),.,1,-,q,a,1,记,t,=,则由,S,5,S,15,S,10,成等差数列得:,S,5,+,S,10,-,2,S,15,=0.,t,(1,-,q,5,+1,-,q,10,-,2+2,q,15,)=0.,即,tq,5,(2,q,10,-,q,5,-,1)=0.,tq,5,0,2,q,10,-,q,5,-,1=0.,下列有两种证法:,法,1:,q,1,可解得,:,q,5,=,-,.,1,2,S,10,2,=,t,2,(1,-,q,10,),2,=,t,2,16,9,2,S,5,(,S,20,-,S,10,)=2,t,2,(1,-,q,5,)(,q,10,-,q,20,)=,t,2,=,S,10,2,.,16,9,2,S,5,S,10,S,20,-,S,10,成等比数列.,法,2:,1+,q,5,=2,q,10,.,S,10,2,S,5,=(1+,q,5,)=,q,10,.,1,-,q,10,2(1,-,q,5,),1,2,S,20,-,S,10,S,10,又,=,-,1=1+,q,10,-,1=,q,10,.,1,-,q,20,1,-,q,10,=.,S,10,2,S,5,S,20,-,S,10,S,10,2,S,5,S,10,S,20,-,S,10,成等比数列.,(2),解:,不一定成立,.,例如,q,=1,时,显然,2,S,5,S,10,S,20,-,S,10,成等比数列,但,S,5,S,15,S,10,不成等差数列.,11.设等比数列 an 的各项为实数,前 n 项的和为Sn,公比为q.(1)若 S5,S15,S10 成等差数列,求证:2S5,S10,S20-S10 成等比数列;(2)若 2S5,S10,S20-S10 成等比数列,试问若 S5,S15,S10一定成等差数列吗?请阐明理由.,12.设数列 an 的前 n 项和为 Sn,若 Sn 是首项为 S1 各项均为正数且公比为q 的等比数列.(1)求数列 an 的通项公式 an(用 S1 和 q 表达);(2)试比较 an+an+2 与 2an+1 的大小,并证明你的结论.,解:,(1),由已知,S,n,=,S,1,q,n,-,1,(,q,0).,当,n,=1,时,a,1,=,S,1,;当,n,2,时,a,n,=,S,n,-,S,n,-,1,=,S,1,(,q,-,1),q,n,-,2,.,(2)当,n,=1,时,a,1,+,a,3,-,2,a,2,=,S,1,+,S,1,(,q,-,1),q,-,2,S,1,(,q,-,1),=,S,1,(,q,2,-,3,q,+3)0.,a,1,+,a,3,2,a,2,;,当,n,2,时,a,n,+,a,n,+2,-,2,a,n,+1,=,S,1,(,q,-,1),q,n,-,2,+,S,1,(,q,-,1),q,n,-,2,S,1,(,q,-,1),q,n,-,1,S,1,0,q,n,-,2,0,当,q,=1,时,(,q,-,1),3,=0,a,n,+,a,n,+2,-,2,a,n,+1,=0,a,n,+,a,n,+2,=2,a,n,+1,;,a,n,=,S,1,(,n,=1),S,1,(,q,-,1),q,n,-,2,(,n,2),=,S,1,(,q,-,1),3,q,n,-,2,.,当,0,q,1,时,(,q,-,1),3,0,a,n,+,a,n,+2,-,2,a,n,+1,0,a,n,+,a,n,+2,1,时,(,q,-,1),3,0,a,n,+,a,n,+2,-,2,a,n,+1,0,a,n,+,a,n,+2,2,a,n,+1,.,总而言之,当 n=1 时,a1+a32a2;,当,n,2,时,若,q,=1,则,a,n,+,a,n,+2,=2,a,n,+1,;,若,0,q,1,则,a,n,+,a,n,+2,1,则,a,n,+,a,n,+2,2,a,n,+1,.,13.下表给出一种“三角形数阵”:已知每一列的数成等差数列,从第三行起每一行的数成等比数列,每一行的公比 都相等.记第 i 行第 j 列的数为 aij(ij,i,jN*).,(1)求,a,83,;,(2)写出 aij 有关 i,j 的体现式;,(3)记第 n 行的和为 An,求数列 An 的前 m 项和 Bm 的体现式;,3,4,3,8,1,2,1,4,16,3,1,4,解:,(1),依题意,a,i,1,成等差数列.,a,11,=,a,21,=,1,2,1,4,每行的公比,q,=.,1,2,a,81,=+(8,-,1),=2.,1,4,1,4,a,31,=,a,32,=,且各行成等比数列,公比都相等,3,8,3,4,1,2,a,83,=2,(),2,=.,1,2,(2)由(1)知,a,i,1,=,+(,i,-,1),=.,1,4,1,4,i,4,i,4,1,2,a,ij,=,a,i,1,(),j,-,1,=,(),j,-,1,=,i,(),j,+1,.,1,2,1,2,(3),A,n,=,a,n,1,1+2,-,1,+2,-,2,+,+2,-,(,n,-,1),=2,-,2,-,(,n,-,1),=,-,n,(),n,+1,.,n,4,n,2,1,2,B,m,=,(1+2+,+,m,),-,(+,+),.,m,2,m,1,2,1,2,2,4,3,8,1,2,设,T,m,=,+,+,+,.,m,2,m,1,2,2,4,3,8,m,+2,2,m,+1,1,2,由错位相减法可求得,T,m,=1,-,m,+2,2,m,+1,B,m,=,+,-,1,.,m,(,m,+1),4,14.设各项均为正数的数列,a,n,和,b,n,满足,5,a,n,5,b,n,5,a,n,+1,成等比数列,lg,b,n,lg,a,n,+1,lg,b,n,+1,成等差数列,且,a,1,=1,b,1,=2,a,2,=3,求通项,a,n,b,n,.,解:,5,a,n,5,b,n,5,a,n,+1,成等比数列,(5,b,n,),2,=5,a,n,5,a,n,+1,2,b,n,=,a,n,+,a,n,+1,.,又,lg,b,n,lg,a,n,+1,lg,b,n,+1,成等差数列,a,n,+1,=,b,n,b,n,+1,.,a,n,=,b,n,-,1,b,n,(,n,2).,2,b,n,=,b,n,b,n,+1,+,b,n,-,1,b,n,(,n,2).,2,b,n,=,b,n,-,1,+,b,n,+1,(,n,2).,又由,lg,b,1,lg,a,2,lg,b,2,成等差数列,且,b,1,=2,a,2,=3,得:,b,2,=.,9,2,b,2,-,b,1,=.,2,2,b,n,是以,2,为首项,为公差的等差数列.,2,2,b,n,=2+(,n,-,1),2,2,2,n,+1,=.,b,n,=.,(,n,+1),2,2,b,n,-,1,=(,n,2).,n,2,2,a,n,=,b,n,-,1,b,n,=(,n,2).,n,(,n,+1),2,又,a,1,=1,亦适合上式,n,(,n,+1),2,a,n,=.,15.,设,a,n,是由正数组成的等比数列,S,n,是其前,n,项和,.(1)证明:,lg,S,n,+lg,S,n,+2,0,使得 =lg(,S,n,+1,-,c,)成立?并证明你的结论.,lg(,S,n,-,c,)+lg(,S,n,+2,-,c,),2,(1),证:,设,等比数列,a,n,的公比为,q,由题设知,a,1,0,q,0.,当,q,=1,时,S,n,=,na,1,S,n,S,n,+2,-,S,n,+1,2,=,na,1,(,n,+2),a,1,-,(,n,+1),2,a,1,2,=,-,a,1,2,0;,当,q,1,时,S,n,=,a,1,(1,-,q,n,),1,-,q,S,n,S,n,+2,-,S,n,+1,2,=,-,a,1,2,(1,-,q,n,)(1,-,q,n,+2,),(1,-,q,),2,a,1,2,(1,-,q,n,+1,),2,(1,-,q,),2,=,-,a,1,2,q,n,0.,S,n,S,n,+2,-,S,n,+1,2,0.,S,n,S,n,+2,S,n,+1,2,.,lg,S,n,S,n,+2,lg,S,n,+1,2,.,lg,S,n,+lg,S,n,+2,2lg,S,n,+1,.,lg,S,n,+lg,S,n,+2,lg,S,n,+1,.,2,=,-,a,1,2,0,使结论成立.,当,q,1,时,(,S,n,-,c,)(,S,n,+2,-,c,),-,(,S,n,+1,-,c,),2,a,1,(1,-,q,n,),1,-,q,=,-,c,-,c,-,-,c,2,a,1,(1,-,q,n,+2,),1,-,q,a,1,(1,-,q,n,+1,),1,-,q,当,q,=1,时,(,S,n,-,c,)(,S,n,+2,-,c,),-,(,S,n,+1,-,c,),2,(,S,n,-,c,)(,S,n,+2,-,c,)=(,S,n,+1,-,c,),2,S,n,-,c,0,解法1:,要使 =lg(,S,n,+1,-,c,)成立,则有,lg(,S,n,-,c,)+lg(,S,n,+2,-,c,),2,(2)是否存在常数,c,0,使得 =lg(,S,n,+1,-,c,)成立?并证明你的结论.,lg(,S,n,-,c,)+lg(,S,n,+2,-,c,),2,=,-,a,1,q,n,a,1,-,c,(1,-,q,),且,a,1,q,n,0,故只能有,a,1,-,c,(1,-,q,)=0,即,c,=,此时,c,0,a,1,0,a,1,1,-,q,0,q,1,.,但当,0,q,1,时,S,n,-,c,=,S,n,-,a,1,1,-,q,a,1,q,n,1,-,q,=,-,0,使结论成立.,故不存在常数,c,0,使得 =lg(,S,n,+1,-,c,),成立.,lg(,S,n,-,c,)+lg(,S,n,+2,-,c,),2,解法2:,假设存在常数,c,0,使 =lg(,S,n,+1,-,c,)成,立,则有,lg(,S,n,-,c,)+lg(,S,n,+2,-,c,),2,(,S,n,-,c)(,S,n,+2,-,c,)=(,S,n,+1,-,c,),2,S,n,-,c,0,S,n,+1,-,c,0,S,n,+2,-,c,0,由,得,S,n,S,n,+2,-,S,n,+1,2,=,c,(,S,n,+,S,n,+2,-,2,S,n,+1,),S,n,+,S,n,+2,-,2,S,n,+1,=(,S,n,-,c,)+(,S,n,+2,-,c,),-,2(,S,n,+1,-,c,),2,(,S,n,-,c,)(,S,n,+2,-,c,),-,2(,S,n,+1,-,c,)=0.,c,0,式右端非负,由(1)知式的左端不大于零,矛盾.,故不存在常数,c,0,使得 =lg(,S,n,+1,-,c,),成立.,lg(,S,n,-,c,)+lg(,S,n,+2,-,c,),2,a,1,=,-,393,a,2,+,a,3,=,-,768,解:,(1)设等差数列,a,n,的公差为,d,前,n,项和为,S,n,.,2,(,-,393)+3,d,=,-,768.,解得,d,=6.,(2)由,(1),知,S,n,=,-,393,n,+3,n,(,n,-,1)=3,n,2,-,396,n,又-,160,b,2,=,-,288.,16.已知,a,n,是等差数列,a,1,=,-,393,a,2,+,a,3,=,-,768,b,n,是公比为,q,(,0,q,1,),的无穷等比数列,b,1,=2,且,b,n,的各项和为,20.(1)写出,a,n,和,b,n,的通项公式;(2)试求满足不等式,-,160,b,2,的正整数,m,.,a,m,+1,+,a,m,+2,+,+,a,2,m,m,+1,a,n,=,-,393+6(,n,-,1)=6,n,-,399.,又由已知,b,1,=2,且,=20,b,1,1,-,q,q,=,.,10,9,b,n,=2(),n,-,1,.,10,9,故,a,n,和,b,n,的通项公式分别为,a,n,=6,n,-,399,b,n,=2(),n,-,1,.,10,9,a,m,+1,+,a,m,+2,+,+,a,2,m,=,S,2,m,-,S,m,=9,m,2,-,396,m,.,原不等式,9,m,2,-,396,m,-,288(,m,+1)(,m,N,*,),m,2,-,12,m,+32,0(,m,N,*,),4,m,8(,m,N,*,).,m,的值为,4,5,6,7,8.,