高中數(shù)學(xué)(人教版B版·必修5)配套練習(xí)2.3等比數(shù)列 第2課時(shí)

672226321012510267222632101251029530等數(shù)列課一、選擇題．在等比數(shù){}，a＋＝，a＋a＝，則＋a等于()n4678A90B．30C．D．[答案D[解析q

245aa()q2089．(2014·慶理，對(duì)任意等比數(shù)列{}下說(shuō)法一定正確的()nAa，，a成比數(shù)列139Ba，，成比數(shù)列2C．，，成比列2D．，，成比數(shù)列369[答案D[解析qaqa3aaD.63936．等比數(shù)列{a}項(xiàng)為正數(shù)，且是和的比中項(xiàng)，則a…a＝n1

)A3

9

B．3C．

11

D．[答案[解析956aaaaa9110935a9110．在等比數(shù){}，aa＝，則的為()n511a11A9B1C．D．[答案D[解析aa3

28291610266642424nm135728291610266642424nm1357qa243aq111．已知等比數(shù)列{}，a＝a，列等差數(shù)列，且＝，b＋等()n379A2C．

B．4D．16[答案[解析aaa≠311777ab{}77bbb8.59．在等比數(shù){}，a>，＝，＋a＝，則等于()nn1414a16C.[答案Aa146[解析

D．a(chǎn)>an1

a3aa2..414a216二、填空題．(2014·蘇，7)在各項(xiàng)均為正數(shù)的等比數(shù){}，＝，a＝a＋a，的是________．n28646[答案[解析aa2

q

2

q

q

q

2

a6qnm＋＋＋a．已知等比數(shù)列{}公q＝，等________．na＋＋＋2468[答案－[解析

aaa135135aaaa24657三、解答題．已知數(shù)列{a}等比數(shù)列．n

22nn4430201510301022nnn2222nn22nn4430201510301022nnn2222nn1n1(1)若a＋＋a＝21，＝216，求a；12123(2)若a＝18a＝72求公比q.354[解析a216a12a36.13aa151aa1531aq3·211a12，a12·()1n

(2)aqa483

3

aq35

1872q

4

4±2.一、選擇題設(shè){}由正數(shù)組成的等比數(shù)列比q2且aa…a＝那a等(n330930

)A2

10

B．2C．

16

D．[答案[解析Aaa…aBaaa…142828Ca…3630qA·2B2CB

10

20

．如果數(shù)列{a}等比數(shù)列，那么)nA數(shù){}等比數(shù)列nB數(shù){a}等數(shù)列nC．{a}等數(shù)列nD．列{na}等數(shù)列n[答案A[解析b

n

a()qnbanna{}2n2n

－a≠n1naannn

ccnann

≠

abacb222abacb222222．在等比數(shù){}，比為q則下列結(jié)論正確的)nA當(dāng)q>1，{}遞增數(shù)列nB當(dāng)0<q<1時(shí){}遞增數(shù)nC．∈N時(shí)，a>0成D．nN時(shí)，a>0成2n[答案[解析28…A11，，…qB1,11482aaDC.135．已知2＝3,2＝6,2＝，則a，，c()A成等差數(shù)列不成等比數(shù)列B成等比數(shù)列不成等差數(shù)列C．等差數(shù)列又成等比數(shù)列D．不等差數(shù)列又不成等比數(shù)列[答案A[解析log3b6log3122cloglog32.2bc36)abA.二、填空題差不為零的等數(shù){}－＋a＝列{}等比數(shù)列b＝ab＝________.n11n[答案[解析2aaaa)a3731174a77ba≠0ba777b68．在3和個(gè)未知數(shù)間填上一個(gè)，使三數(shù)成等差數(shù)列，若中間項(xiàng)減去6則等數(shù)列，則此未知數(shù)是．[答案3或27[解析

ab

24444448322nn522222224444448322nn52222222227.三、解答題．{a}等比數(shù)列，且a＝，a＋a＝20，求n19711[解析{}naaa17aatt3a16a3aaaq33731q5q4.a16aa(1q20311q，q.4aa6411．{}各項(xiàng)均為正數(shù)的等比數(shù)列b＝log，b＋b＋b＝，bb＝－，求此等比數(shù)列的nn122通項(xiàng)公式.n[解析bb31log(a)323aa813aaa2bb2122{}()·log(2)3.n21)3logq±2.22q，{}nqn2

3

a2n

2n

．(2013·國(guó)大綱理，17)等差數(shù)列{}前項(xiàng)為，知＝nn{}通公式．n[解析{}nSa