重慶高三數(shù)列試題及答案

一、單項選擇題（每題2分，共10題）1.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(d=2\)，則\(a_{5}\)等于（）A.7B.9C.11D.132.等比數(shù)列\(zhòng)(\{b_{n}\}\)中，\(b_{1}=2\)，公比\(q=3\)，則\(b_{3}\)是（）A.18B.12C.6D.273.數(shù)列\(zhòng)(\{a_{n}\}\)通項公式\(a_{n}=2n-1\)，則\(a_{3}\)的值為（）A.5B.4C.3D.24.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}\)，若\(a_{3}=5\)，則\(S_{5}\)為（）A.15B.20C.25D.305.等比數(shù)列\(zhòng)(\{b_{n}\}\)中，若\(b_{2}b_{4}=9\)，則\(b_{3}\)等于（）A.3B.-3C.\(\pm3\)D.\(\pm9\)6.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=a_{n}+3\)，\(a_{1}=2\)，則\(a_{4}\)為（）A.11B.9C.7D.57.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}\)，則\(a_{2}\)的值是（）A.3B.4C.5D.68.等比數(shù)列\(zhòng)(\{b_{n}\}\)中，\(b_{1}=1\)，\(b_{4}=8\)，則公比\(q\)為（）A.2B.-2C.3D.49.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}+a_{5}=10\)，則\(a_{3}\)的值為（）A.5B.8C.10D.1210.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n}=3^{n}\)，則\(a_{2}\)與\(a_{1}\)的比值為（）A.2B.3C.4D.9二、多項選擇題（每題2分，共10題）1.以下關(guān)于等差數(shù)列性質(zhì)正確的有（）A.若\(m+n=p+q\)，則\(a_{m}+a_{n}=a_{p}+a_{q}\)B.公差\(d\)可以為0C.前\(n\)項和\(S_{n}=\frac{n(a_{1}+a_{n})}{2}\)D.\(a_{n}=a_{1}+(n-1)d\)2.等比數(shù)列\(zhòng)(\{b_{n}\}\)中，可能成立的有（）A.\(b_{1}b_{3}=b_{2}^{2}\)B.公比\(q\)可以為負(fù)數(shù)C.\(b_{n}=b_{1}q^{n-1}\)D.若\(m+n=p+q\)，則\(b_{m}b_{n}=b_{p}b_{q}\)3.下列數(shù)列是等差數(shù)列的有（）A.\(1,3,5,7,\cdots\)B.\(2,4,8,16,\cdots\)C.\(5,5,5,5,\cdots\)D.\(-1,-3,-5,-7,\cdots\)4.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)，其公差\(d\)不為\(0\)，那么（）A.數(shù)列\(zhòng)(\{a_{n}\}\)單調(diào)遞增（\(d\gt0\)）B.數(shù)列\(zhòng)(\{a_{n}\}\)可能是常數(shù)列C.\(a_{n+1}-a_{n}=d\)D.\(a_{n}=a_{m}+(n-m)d\)5.等比數(shù)列\(zhòng)(\{b_{n}\}\)的公比為\(q\)，則（）A.當(dāng)\(q\gt1\)時，數(shù)列\(zhòng)(\{b_{n}\}\)單調(diào)遞增B.\(b_{n+1}=b_{n}q\)C.當(dāng)\(q\lt0\)時，數(shù)列\(zhòng)(\{b_{n}\}\)的項正負(fù)交替出現(xiàn)D.若\(b_{1}=1\)，\(q=2\)，則\(b_{4}=8\)6.以下能確定數(shù)列\(zhòng)(\{a_{n}\}\)為等差數(shù)列的條件是（）A.\(a_{n+1}-a_{n}=2\)（\(n\inN^\)）B.\(2a_{n}=a_{n-1}+a_{n+1}\)（\(n\geqslant2\)）C.\(a_{1}=1\)，\(a_{2}=3\)，\(a_{3}=5\)D.\(a_{n}=2n+1\)7.關(guān)于等比數(shù)列前\(n\)項和\(S_{n}\)，正確的有（）A.當(dāng)\(q=1\)時，\(S_{n}=nb_{1}\)B.\(S_{n}=\frac{b_{1}(1-q^{n})}{1-q}(q\neq1)\)C.\(S_{n}\)在\(q\neq1\)時，與\(n\)是指數(shù)函數(shù)關(guān)系D.\(S_{n}\)一定能寫成\(Aq^{n}+B\)（\(A\)、\(B\)為常數(shù)）形式8.已知數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列，\(S_{n}\)為其前\(n\)項和，下列說法正確的是（）A.\(S_{n}\)，\(S_{2n}-S_{n}\)，\(S_{3n}-S_{2n}\)仍成等差數(shù)列B.\(a_{m}+a_{n+1}=a_{m+1}+a_{n}\)C.若\(a_{1}\gt0\)，\(d\lt0\)，則\(S_{n}\)有最大值D.若\(a_{1}\lt0\)，\(d\gt0\)，則\(S_{n}\)有最小值9.下列各項中，是等比數(shù)列的有（）A.\(1,\frac{1}{2},\frac{1}{4},\frac{1}{8},\cdots\)B.\(1,-1,1,-1,\cdots\)C.\(0,0,0,0,\cdots\)D.\(2,4,8,16,\cdots\)10.對于等差數(shù)列\(zhòng)(\{a_{n}\}\)與等比數(shù)列\(zhòng)(\{b_{n}\}\)，下列說法中正確的有（）A.若\(\{a_{n}\}\)各項為正，則\(\{\sqrt{a_{n}}\}\)可能成等差數(shù)列B.若\(\{b_{n}\}\)各項為正，則\(\{\lgb_{n}\}\)可能成等差數(shù)列C.若\(\{a_{n}\}\)成等差數(shù)列，則\(\{2^{a_{n}}\}\)可能成等比數(shù)列D.若\(\{b_{n}\}\)成等比數(shù)列，則\(\{\sqrt{b_{n}}\}\)可能成等比數(shù)列三、判斷題（每題2分，共10題）1.數(shù)列\(zhòng)(1,2,3,4,5\)是等差數(shù)列。（）2.若數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=2a_{n}\)，則\(\{a_{n}\}\)是等比數(shù)列。（）3.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}\)一定是關(guān)于\(n\)的二次函數(shù)。（）4.等比數(shù)列\(zhòng)(\{b_{n}\}\)中，若\(b_{1}\lt0\)，\(q\gt1\)，則數(shù)列\(zhòng)(\{b_{n}\}\)單調(diào)遞減。（）5.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}+1\)，則\(\{a_{n}\}\)是等差數(shù)列。（）6.等比數(shù)列\(zhòng)(\{b_{n}\}\)的公比\(q\)不能為\(0\)。（）7.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{n}=a_{1}+nd\)。（）8.若\(a,b,c\)成等比數(shù)列，則\(b^{2}=ac\)。（）9.數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}=1\)，\(a_{n+1}-a_{n}=2\)，則\(a_{5}=9\)。（）10.等比數(shù)列\(zhòng)(\{b_{n}\}\)中，\(S_{2}\)，\(S_{4}-S_{2}\)，\(S_{6}-S_{4}\)也成等比數(shù)列。（）四、簡答題（每題5分，共4題）1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=3\)，\(d=2\)，求\(a_{n}\)的通項公式。答案：根據(jù)等差數(shù)列通項公式\(a_{n}=a_{1}+(n-1)d\)，將\(a_{1}=3\)，\(d=2\)代入，可得\(a_{n}=3+2(n-1)=2n+1\)。2.等比數(shù)列\(zhòng)(\{b_{n}\}\)中，\(b_{1}=2\)，\(b_{3}=8\)，求公比\(q\)。答案：由等比數(shù)列通項公式\(b_{n}=b_{1}q^{n-1}\)，則\(b_{3}=b_{1}q^{2}\)，把\(b_{1}=2\)，\(b_{3}=8\)代入，得\(8=2q^{2}\)，即\(q^{2}=4\)，解得\(q=\pm2\)。3.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}-2n\)，求\(a_{3}\)的值。答案：\(a_{3}=S_{3}-S_{2}\)，\(S_{2}=2^{2}-2\times2=0\)，\(S_{3}=3^{2}-2\times3=3\)，所以\(a_{3}=3-0=3\)。4.等比數(shù)列\(zhòng)(\{b_{n}\}\)中，\(b_{1}=1\)，公比\(q=3\)，求前\(n\)項和\(S_{n}\)的表達(dá)式。答案：當(dāng)\(q=3\neq1\)時，根據(jù)等比數(shù)列前\(n\)項和公式\(S_{n}=\frac{b_{1}(1-q^{n})}{1-q}\)，將\(b_{1}=1\)，\(q=3\)代入，可得\(S_{n}=\frac{1-3^{n}}{1-3}=\frac{3^{n}-1}{2}\)。五、討論題（每題5分，共4題）1.討論等差數(shù)列和等比數(shù)列在實(shí)際生活中的應(yīng)用實(shí)例。答案：等差數(shù)列如在每月等額還款的房貸中，每月還款中本金與利息總和構(gòu)成等差數(shù)列；等比數(shù)列像細(xì)胞分裂，每經(jīng)過一定時間細(xì)胞數(shù)量以等比形式增長。它們在經(jīng)濟(jì)、科學(xué)研究等多方面有重要應(yīng)用。2.分析判斷一個數(shù)列是等差數(shù)列或等比數(shù)列有哪些方法？答案：判斷等差數(shù)列可利用定義\(a_{n+1}-a_{n}\)為常數(shù)、中項性質(zhì)\(2a_{n}=a_{n-1}+a_{n+1}\)；判斷等比數(shù)列可用定義\(\frac{a_{n+1}}{a_{n}}\)為非零常數(shù)、中項性質(zhì)\(a_{n}^{2}=a_{n-1}a_{n+1}\)，還可看通項公式與前\(n\)項和公式的特征。3.若一個數(shù)列既是等差數(shù)列又是等比數(shù)列，這個數(shù)列有什么特點(diǎn)？答案：這樣的數(shù)列一定是非