新高二數(shù)列測試題及答案

一、單項選擇題（每題2分，共10題）1.數(shù)列\(zhòng)(1,3,5,7,9\)的通項公式是（）A.\(a_{n}=2n-1\)B.\(a_{n}=n+1\)C.\(a_{n}=2n+1\)D.\(a_{n}=n-1\)2.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(d=3\)，則\(a_{5}\)的值為（）A.\(14\)B.\(15\)C.\(16\)D.\(17\)3.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(q=2\)，則\(a_{4}\)等于（）A.\(4\)B.\(8\)C.\(16\)D.\(32\)4.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}\)，則\(a_{3}\)的值為（）A.\(5\)B.\(6\)C.\(7\)D.\(8\)5.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}+a_{5}=10\)，則\(a_{4}\)的值為（）A.\(5\)B.\(6\)C.\(7\)D.\(8\)6.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}a_{6}=16\)，則\(a_{4}\)的值為（）A.\(4\)B.\(-4\)C.\(\pm4\)D.\(8\)7.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=a_{n}+2\)，\(a_{1}=1\)，則\(a_{5}\)為（）A.\(9\)B.\(10\)C.\(11\)D.\(12\)8.若\(a\)，\(b\)，\(c\)成等差數(shù)列，則\(2b\)等于（）A.\(a+c\)B.\(ac\)C.\(a-c\)D.\(\frac{a}{c}\)9.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{3}=8\)，則公比\(q\)為（）A.\(2\)B.\(-2\)C.\(\pm2\)D.\(4\)10.數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=3n-2\)，則\(a_{10}\)的值是（）A.\(28\)B.\(29\)C.\(30\)D.\(31\)二、多項選擇題（每題2分，共10題）1.以下哪些是等差數(shù)列（）A.\(1,3,5,7\)B.\(2,4,8,16\)C.\(5,5,5,5\)D.\(1,2,4,7\)2.等比數(shù)列的性質(zhì)正確的有（）A.\(a_{m}a_{n}=a_{p}a_{q}\)（\(m+n=p+q\)）B.\(a_{n}=a_{1}q^{n-1}\)C.\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)（\(q\neq1\)）D.\(S_{n}=na_{1}\)（\(q=1\)）3.已知數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列，\(a_{3}=5\)，\(a_{5}=9\)，則（）A.公差\(d=2\)B.\(a_{1}=1\)C.\(a_{7}=13\)D.\(a_{4}=7\)4.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=2\)，\(a_{1}=1\)，則（）A.\(a_{n}=2n-1\)B.\(a_{5}=9\)C.前\(n\)項和\(S_{n}=n^{2}\)D.該數(shù)列是等差數(shù)列5.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{3}=4\)，則（）A.公比\(q=2\)B.公比\(q=-2\)C.\(a_{2}=2\)D.\(a_{2}=-2\)6.下列關(guān)于數(shù)列的說法正確的是（）A.常數(shù)列既是等差數(shù)列也是等比數(shù)列（非零常數(shù)列）B.等差數(shù)列的通項公式是關(guān)于\(n\)的一次函數(shù)C.等比數(shù)列的前\(n\)項和公式有兩種形式D.數(shù)列的通項公式可以唯一確定一個數(shù)列7.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}+a_{5}=10\)，則（）A.\(a_{3}=5\)B.\(a_{2}+a_{4}=10\)C.若\(d=1\)，則\(a_{n}=n+2\)D.前\(5\)項和\(S_{5}=25\)8.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{4}=8\)，則（）A.公比\(q=2\)B.\(a_{1}=1\)C.\(a_{3}=4\)D.前\(3\)項和\(S_{3}=7\)9.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}+n\)，則（）A.\(a_{1}=2\)B.\(a_{n}=2n\)C.數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列D.\(a_{2}=4\)10.已知數(shù)列\(zhòng)(\{a_{n}\}\)是等比數(shù)列，\(a_{3}=9\)，\(a_{6}=243\)，則（）A.公比\(q=3\)B.\(a_{1}=1\)C.\(a_{4}=27\)D.前\(5\)項和\(S_{5}=121\)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.等比數(shù)列的公比\(q\)可以為\(0\)。（）3.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(m+n=p+q\)，則\(a_{m}+a_{n}=a_{p}+a_{q}\)。（）4.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{n}=a_{m}q^{n-m}\)。（）5.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=3n\)，則\(a_{n}=3\)。（）6.常數(shù)列一定是等差數(shù)列。（）7.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(q=-1\)，則\(a_{2}=-1\)。（）8.等差數(shù)列的公差\(d\)可以是負(fù)數(shù)。（）9.若\(a\)，\(b\)，\(c\)成等比數(shù)列，則\(b^{2}=ac\)。（）10.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=2a_{n}\)，\(a_{1}=1\)，則\(a_{n}=2^{n-1}\)。（）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\(a_{n}\)。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，得\(a_{n}=3+2(n-1)=2n+1\)。2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(q=3\)，求前\(n\)項和\(S_{n}\)。答案：由等比數(shù)列前\(n\)項和公式\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)（\(q\neq1\)），把\(a_{1}=2\)，\(q=3\)代入，得\(S_{n}=\frac{2(1-3^{n})}{1-3}=3^{n}-1\)。3.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}-n\)，求\(a_{n}\)。答案：當(dāng)\(n=1\)時，\(a_{1}=S_{1}=1^{2}-1=0\)；當(dāng)\(n\geq2\)時，\(a_{n}=S_{n}-S_{n-1}=(n^{2}-n)-[(n-1)^{2}-(n-1)]=2n-2\)。\(n=1\)時也滿足，所以\(a_{n}=2n-2\)。4.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=7\)，\(a_{5}=11\)，求\(a_{1}\)和\(d\)。答案：因為\(a_{5}-a_{3}=2d\)，所以\(2d=11-7=4\)，得\(d=2\)。又\(a_{3}=a_{1}+2d\)，即\(7=a_{1}+2×2\)，解得\(a_{1}=3\)。五、討論題（每題5分，共4題）1.討論等差數(shù)列和等比數(shù)列在實際生活中的應(yīng)用。答案：等差數(shù)列在計算定期等額存款利息、樓層高度變化等方面有應(yīng)用；等比數(shù)列常用于計算復(fù)利、細胞分裂、病毒傳播等問題，通過通項公式和求和公式能解決相關(guān)實際數(shù)量變化的計算。2.如何判斷一個數(shù)列是等差數(shù)列還是等比數(shù)列？答案：判斷等差數(shù)列可看相鄰兩項的差是否為常數(shù)，即\(a_{n+1}-a_{n}\)是否為定值；判斷等比數(shù)列看相鄰兩項的比是否為常數(shù)，即\(\frac{a_{n+1}}{a_{n}}\)是否為定值（\(a_{n}\neq0\)）。3.數(shù)列的通項公式和前\(n\)項和公式有什么關(guān)系？答案：已知通項公式\(a_{n}\)，可通過\(S_{n}=a_{1}+a_{2}+\cdots+a_{n}\)求前\(n\)項和；已知\(S_{n}\)，\(n=1\)時，\(a_{1}=S_{1}\)，\(n\geq2\)時，\(a_{n}=S_{n}-S_{n-1}\)，二者相互關(guān)聯(lián)用于研究數(shù)列性質(zhì)。4.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？答案：非零常數(shù)列既是等差數(shù)列（公差為\(0\)）又是等比數(shù)列（公比為\(1\)）。因為等差數(shù)列\(zhòng)(a_{n+1}-a_{n}=d\)，等比數(shù)列\(zhòng)(\frac{a_{n+1}}{a_{n}}=q\)，要同時滿足，只能\(d=0\)且\(q=1\)，即數(shù)列各項都相等且不為\(0\)。