高考天津數(shù)列試題及答案

單項(xiàng)選擇題（每題2分，共10題）1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{3}=5\)，則公差\(d=(\)\)A.1B.2C.3D.42.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{5}=16\)，則公比\(q=(\)\)A.-2B.2C.4D.-43.數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=2n-1\)，則\(a_{5}=(\)\)A.7B.8C.9D.104.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}\)，若\(a_{3}=3\)，\(S_{5}=(\)\)A.10B.15C.20D.255.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{4}=8\)，則\(a_{3}=(\)\)A.4B.-4C.2D.-26.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=a_{n}+2\)，\(a_{1}=1\)，則\(a_{4}=(\)\)A.5B.6C.7D.87.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}+a_{4}=10\)，則\(a_{3}=(\)\)A.5B.6C.7D.88.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}a_{5}=16\)，則\(a_{4}=(\)\)A.4B.-4C.\(\pm4\)D.89.數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=n^{2}-n\)，則\(a_{3}-a_{2}=(\)\)A.4B.5C.6D.710.等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差\(d=2\)，\(a_{1}=1\)，則\(a_{n}=(\)\)A.\(2n-1\)B.\(2n+1\)C.\(n+1\)D.\(n-1\)多項(xiàng)選擇題（每題2分，共10題）1.以下關(guān)于等差數(shù)列的說(shuō)法正確的是（）A.若\(\{a_{n}\}\)是等差數(shù)列，則\(a_{n+1}-a_{n}\)為常數(shù)B.等差數(shù)列的通項(xiàng)公式是關(guān)于\(n\)的一次函數(shù)C.若\(a,b,c\)成等差數(shù)列，則\(2b=a+c\)D.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=na_{1}+\frac{n(n-1)}{2}d\)2.對(duì)于等比數(shù)列\(zhòng)(\{a_{n}\}\)，下列說(shuō)法正確的是（）A.公比\(q\neq0\)B.若\(a_{1},a_{3},a_{5}\)成等比數(shù)列，則\(a_{3}^{2}=a_{1}a_{5}\)C.等比數(shù)列的通項(xiàng)公式\(a_{n}=a_{1}q^{n-1}\)D.等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}(q\neq1)\)3.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=2\)，\(a_{1}=1\)，則（）A.\(\{a_{n}\}\)是等差數(shù)列B.\(a_{n}=2n-1\)C.\(S_{n}=n^{2}\)D.\(a_{5}=9\)4.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{3}=4\)，則（）A.\(q=2\)B.\(q=-2\)C.\(a_{2}=2\)D.\(a_{2}=-2\)5.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(d=2\)，則（）A.\(a_{n}=2n-1\)B.\(S_{n}=n^{2}\)C.\(a_{5}=9\)D.\(a_{2}+a_{3}=8\)6.以下數(shù)列中，可能是等比數(shù)列的有（）A.\(1,2,4,8,\cdots\)B.\(-1,1,-1,1,\cdots\)C.\(0,0,0,0,\cdots\)D.\(2,2,2,2,\cdots\)7.關(guān)于數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=n(n+1)\)，下列說(shuō)法正確的是（）A.\(a_{1}=2\)B.\(a_{2}=6\)C.\(a_{n}-a_{n-1}=2n\)D.該數(shù)列不是等差數(shù)列8.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}\)，則（）A.\(a_{1}=1\)B.\(a_{2}=3\)C.\(d=2\)D.\(a_{n}=2n-1\)9.等比數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{2}a_{4}=16\)，則（）A.\(q=2\)B.\(q=-2\)C.\(a_{3}=4\)D.\(a_{3}=-4\)10.數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{n+1}=2a_{n}\)，則（）A.\(\{a_{n}\}\)是等比數(shù)列B.\(a_{n}=2^{n-1}\)C.\(S_{n}=2^{n}-1\)D.\(a_{3}=4\)判斷題（每題2分，共10題）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.若\(\{a_{n}\}\)是等差數(shù)列，\(m+n=p+q\)，則\(a_{m}+a_{n}=a_{p}+a_{q}\)。（）4.若\(\{a_{n}\}\)是等比數(shù)列，\(m+n=p+q\)，則\(a_{m}a_{n}=a_{p}a_{q}\)。（）5.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}\)一定是關(guān)于\(n\)的二次函數(shù)。（）6.等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}\)一定不能為\(0\)。（）7.若數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=3n+2\)，則\(\{a_{n}\}\)是等差數(shù)列。（）8.若數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=2^{n}\)，則\(\{a_{n}\}\)是等比數(shù)列。（）9.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}\gt0\)，\(d\lt0\)，則\(S_{n}\)有最大值。（）10.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}\gt0\)，\(0\ltq\lt1\)，則\(a_{n}\)單調(diào)遞減。（）簡(jiǎn)答題（每題5分，共4題）1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(d=3\)，求\(a_{n}\)與\(S_{n}\)。答案：\(a_{n}=a_{1}+(n-1)d=2+3(n-1)=3n-1\)；\(S_{n}=na_{1}+\frac{n(n-1)}{2}d=2n+\frac{3n(n-1)}{2}=\frac{3n^{2}+n}{2}\)。2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(q=2\)，求\(a_{5}\)和\(S_{5}\)。答案：\(a_{n}=a_{1}q^{n-1}\)，所以\(a_{5}=1\times2^{4}=16\)；\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)，\(S_{5}=\frac{1\times(1-2^{5})}{1-2}=31\)。3.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}\)，求\(a_{n}\)。答案：當(dāng)\(n=1\)時(shí)，\(a_{1}=S_{1}=1\)；當(dāng)\(n\geq2\)時(shí)，\(a_{n}=S_{n}-S_{n-1}=n^{2}-(n-1)^{2}=2n-1\)，\(n=1\)時(shí)也滿足，所以\(a_{n}=2n-1\)。4.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{7}=13\)，求\(a_{1}\)與\(d\)。答案：由\(a_{n}=a_{1}+(n-1)d\)可得\(\begin{cases}a_{1}+2d=5\\a_{1}+6d=13\end{cases}\)，兩式相減得\(4d=8\)，\(d=2\)，代入\(a_{1}+2d=5\)得\(a_{1}=1\)。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，舉例說(shuō)明。答案：等差數(shù)列如按固定金額定期存款，每月存入金額相同，存款總數(shù)成等差數(shù)列；等比數(shù)列如細(xì)胞分裂，每次分裂數(shù)量是上次的固定倍數(shù)，細(xì)胞總數(shù)成等比數(shù)列。它們可用于計(jì)算經(jīng)濟(jì)增長(zhǎng)、資源消耗等實(shí)際問(wèn)題。2.分析在已知數(shù)列前\(n\)項(xiàng)和\(S_{n}\)求通項(xiàng)公式\(a_{n}\)時(shí)，需要注意什么？答案：要注意分\(n=1\)和\(n\geq2\)兩種情況。\(n=1\)時(shí)，\(a_{1}=S_{1}\)；\(n\geq2\)時(shí)，\(a_{n}=S_{n}-S_{n-1}\)，最后要檢驗(yàn)\(n=1\)時(shí)的結(jié)果是否滿足\(n\geq2\)時(shí)的通項(xiàng)公式。3.探討如何判斷一個(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)列可看相鄰兩項(xiàng)差是否為常數(shù)，即\(a_{n+1}-a