中職數(shù)列試題及答案解析

一、單項(xiàng)選擇題（每題2分，共20分）1.數(shù)列\(zhòng)(2,4,6,8,10\cdots\)的通項(xiàng)公式是（）A.\(a_{n}=2n-1\)B.\(a_{n}=2n\)C.\(a_{n}=n+1\)D.\(a_{n}=n^{2}\)2.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(d=3\)，則\(a_{5}\)的值為（）A.14B.15C.16D.173.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(q=2\)，則\(a_{4}\)等于（）A.4B.8C.16D.324.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}\)，則\(a_{3}\)的值為（）A.5B.6C.7D.85.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}+a_{5}=10\)，則\(a_{4}\)的值為（）A.5B.6C.7D.86.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{4}=8\)，則公比\(q\)為（）A.2B.-2C.\(\pm2\)D.47.數(shù)列\(zhòng)(1,-1,1,-1\cdots\)的一個(gè)通項(xiàng)公式是（）A.\(a_{n}=(-1)^{n}\)B.\(a_{n}=(-1)^{n+1}\)C.\(a_{n}=(-1)^{n-1}\)D.\(a_{n}=(-1)^{n+2}\)8.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和公式\(S_{n}=\)（）A.\(na_{1}+\frac{n(n-1)}{2}d\)B.\(na_{1}+nd\)C.\(\frac{n(a_{1}+a_{n})}{2}\)D.\(n(a_{1}+a_{n})\)9.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=3\)，\(a_{3}=12\)，則\(a_{2}\)的值為（）A.6B.-6C.\(\pm6\)D.910.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=a_{n}+2\)，\(a_{1}=1\)，則\(a_{5}\)的值為（）A.7B.9C.11D.13答案：1.B2.C3.B4.A5.A6.C7.B8.C9.C10.B二、多項(xiàng)選擇題（每題2分，共20分）1.以下哪些是等差數(shù)列（）A.\(1,3,5,7\cdots\)B.\(2,4,8,16\cdots\)C.\(5,5,5,5\cdots\)D.\(1,\frac{3}{2},2,\frac{5}{2}\cdots\)2.等比數(shù)列的通項(xiàng)公式為\(a_{n}=a_{1}q^{n-1}\)，其中（）A.\(a_{1}\)是首項(xiàng)B.\(q\)是公比C.\(n\)是項(xiàng)數(shù)D.\(a_{n}\)是第\(n\)項(xiàng)3.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，下列說(shuō)法正確的是（）A.若\(m+n=p+q\)，則\(a_{m}+a_{n}=a_{p}+a_{q}\)B.\(a_{n}=a_{1}+(n-1)d\)C.公差\(d\)可以為\(0\)D.前\(n\)項(xiàng)和\(S_{n}\)一定是關(guān)于\(n\)的二次函數(shù)4.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，下列說(shuō)法正確的是（）A.若\(m+n=p+q\)，則\(a_{m}\cdota_{n}=a_{p}\cdota_{q}\)B.\(a_{n}=a_{1}q^{n-1}\)C.公比\(q\)不能為\(0\)D.所有項(xiàng)都不能為\(0\)5.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}\)與\(a_{n}\)的關(guān)系是（）A.\(a_{1}=S_{1}\)B.\(a_{n}=S_{n}-S_{n-1}(n\geq2)\)C.\(S_{n}=a_{1}+a_{2}+\cdots+a_{n}\)D.\(a_{n}\)與\(S_{n}\)沒(méi)有直接關(guān)系6.以下數(shù)列中，哪些是遞增數(shù)列（）A.\(a_{n}=2n+1\)B.\(a_{n}=-n+5\)C.\(a_{n}=3^{n}\)D.\(a_{n}=(\frac{1}{2})^{n}\)7.等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差\(d\lt0\)，則該數(shù)列（）A.是遞減數(shù)列B.是遞增數(shù)列C.先增后減D.可能有最大值8.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}\lt0\)，\(0\ltq\lt1\)，則該數(shù)列（）A.是遞增數(shù)列B.是遞減數(shù)列C.所有項(xiàng)都為負(fù)D.所有項(xiàng)都為正9.已知數(shù)列\(zhòng)(\{a_{n}\}\)為等差數(shù)列，\(a_{3}=5\)，\(a_{7}=13\)，則（）A.公差\(d=2\)B.\(a_{1}=1\)C.\(a_{5}=9\)D.\(S_{5}=25\)10.已知數(shù)列\(zhòng)(\{a_{n}\}\)為等比數(shù)列，\(a_{2}=4\)，\(a_{4}=16\)，則（）A.公比\(q=2\)B.\(a_{1}=2\)C.\(a_{3}=8\)D.\(S_{3}=14\)答案：1.ACD2.ABCD3.ABC4.ABCD5.ABC6.AC7.AD8.BC9.ABCD10.ABC三、判斷題（每題2分，共20分）1.常數(shù)列一定是等差數(shù)列。（）2.常數(shù)列一定是等比數(shù)列。（）3.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}\gt0\)，\(d\lt0\)，則數(shù)列有最大值。（）4.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}\gt0\)，\(q\gt1\)，則數(shù)列是遞增數(shù)列。（）5.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}+1\)，則\(a_{n}=2n-1\)。（）6.若\(a,b,c\)成等差數(shù)列，則\(2b=a+c\)。（）7.若\(a,b,c\)成等比數(shù)列，則\(b^{2}=ac\)。（）8.等差數(shù)列的通項(xiàng)公式\(a_{n}\)是關(guān)于\(n\)的一次函數(shù)。（）9.等比數(shù)列的前\(n\)項(xiàng)和\(S_{n}\)不可能為\(0\)。（）10.數(shù)列\(zhòng)(1,2,3,4,1,2,3,4\cdots\)是周期數(shù)列。（）答案：1.√2.×3.√4.√5.×6.√7.√8.×9.×10.√四、簡(jiǎn)答題（每題5分，共20分）1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=3\)，\(d=2\)，求\(a_{n}\)。答案：根據(jù)等差數(shù)列通項(xiàng)公式\(a_{n}=a_{1}+(n-1)d\)，將\(a_{1}=3\)，\(d=2\)代入，得\(a_{n}=3+2(n-1)=2n+1\)。2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{3}=8\)，求公比\(q\)。答案：由等比數(shù)列通項(xiàng)公式\(a_{n}=a_{1}q^{n-1}\)，\(a_{3}=a_{1}q^{2}\)，將\(a_{1}=2\)，\(a_{3}=8\)代入得\(8=2q^{2}\)，即\(q^{2}=4\)，所以\(q=\pm2\)。3.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}-n\)，求\(a_{n}\)。答案：當(dāng)\(n=1\)時(shí)，\(a_{1}=S_{1}=1^{2}-1=0\)；當(dāng)\(n\geq2\)時(shí)，\(a_{n}=S_{n}-S_{n-1}=n^{2}-n-[(n-1)^{2}-(n-1)]=2n-2\)。\(n=1\)時(shí)也滿足，所以\(a_{n}=2n-2\)。4.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{5}=10\)，\(a_{12}=31\)，求\(a_{1}\)和\(d\)。答案：由等差數(shù)列通項(xiàng)公式\(a_{n}=a_{1}+(n-1)d\)，可得\(\begin{cases}a_{1}+4d=10\\a_{1}+11d=31\end{cases}\)，兩式相減得\(7d=21\)，\(d=3\)，將\(d=3\)代入\(a_{1}+4d=10\)，得\(a_{1}=10-12=-2\)。五、討論題（每題5分，共20分）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)列如銀行零存整取，每月固定存相同金額，本息和按等差數(shù)列增長(zhǎng)。等比數(shù)列如細(xì)胞分裂，每經(jīng)過(guò)一定時(shí)間細(xì)胞數(shù)量按等比數(shù)列增加。在經(jīng)濟(jì)、生物等領(lǐng)域都有廣泛應(yīng)用。2.如何判斷一個(gè)數(shù)列是等差數(shù)列還是等比數(shù)列？答案：判斷等差數(shù)列，看相鄰兩項(xiàng)差是否為常數(shù)，即\(a_{n+1}-a_{n}=d\)（\(d\)為常數(shù)）；判斷等比數(shù)列，看相鄰兩項(xiàng)比值是否為常數(shù)，即\(\frac{a_{n+1}}{a_{n}}=q\)（\(q\)為非零常數(shù)）。3.數(shù)列的通項(xiàng)公式和前\(n\)項(xiàng)和公式有什么聯(lián)系？答案：\(a_{1}=S_{1}\)