高考數(shù)列真題及答案

一、單項選擇題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(a_{3}=5\)，\(S_{9}=81\)，則\(a_{7}=\)（）A.18B.13C.9D.7答案：B2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{5}=16\)，則公比\(q\)為（）A.2B.3C.4D.8答案：A3.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}=2a_{n}+1\)，則\(a_{3}\)的值為（）A.7B.9C.11D.13答案：A4.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差\(d\neq0\)，且\(a_{1}\)，\(a_{3}\)，\(a_{9}\)成等比數(shù)列，則\(\frac{a_{1}+a_{3}+a_{9}}{a_{2}+a_{4}+a_{10}}\)的值是（）A.\(\frac{13}{16}\)B.\(\frac{15}{16}\)C.\(\frac{11}{16}\)D.\(\frac{12}{16}\)答案：A5.若數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式為\(a_{n}=2n-1\)，\(b_{n}=(-1)^{n-1}\frac{4n}{a_{n}a_{n+1}}\)，則數(shù)列\(zhòng)(\{b_{n}\}\)的前\(n\)項和\(T_{n}\)為（）A.\(\frac{2n}{2n+1}\)B.\(\frac{4n}{2n+1}\)C.\(\frac{2n}{2n-1}\)D.\(\frac{4n}{2n-1}\)答案：B6.已知數(shù)列\(zhòng)(\{a_{n}\}\)是等比數(shù)列，\(S_{n}\)是其前\(n\)項和，若\(a_{2}a_{3}=2a_{1}\)，且\(a_{4}\)與\(2a_{7}\)的等差中項為\(\frac{5}{4}\)，則\(S_{5}=\)（）A.35B.33C.31D.29答案：C7.數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{n+1}=a_{n}+\frac{1}{n(n+1)}\)，則\(a_{10}\)的值為（）A.\(\frac{9}{10}\)B.\(\frac{19}{10}\)C.\(\frac{17}{10}\)D.\(\frac{11}{10}\)答案：B8.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(S_{17}=170\)，則\(a_{9}\)的值為（）A.10B.15C.20D.25答案：A9.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}+a_{2}+a_{3}=1\)，\(a_{4}+a_{5}+a_{6}=-2\)，則該等比數(shù)列的公比\(q\)為（）A.-1B.-2C.-\sqrt{2}D.\sqrt[3]{-2}答案：D10.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=\frac{1}{1-a_{n}}\)，\(a_{8}=2\)，則\(a_{1}\)等于（）A.\(\frac{1}{2}\)B.\(\frac{1}{3}\)C.\(\frac{1}{4}\)D.\(\frac{1}{5}\)答案：A二、多項選擇題1.設(shè)等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)，前\(n\)項和為\(S_{n}\)，若\(a_{3}=12\)，\(S_{12}\gt0\)，\(S_{13}\lt0\)，則下列結(jié)論正確的是（）A.數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列B.\(S_{5}=60\)C.\(-\frac{24}{7}\ltd\lt-3\)D.\(S_{1}\)，\(S_{2}\)，\(\cdots\)，\(S_{12}\)中最大的是\(S_{6}\)答案：BCD2.已知等比數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=\frac{1}{4}\)，\(a_{3}a_{5}=4(a_{4}-1)\)，則下列結(jié)論正確的是（）A.\(a_{2}=\frac{1}{2}\)B.\(a_{n}=2^{n-3}\)C.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=2^{n-2}-\frac{1}{4}\)D.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項積\(T_{n}\)的最大值為\(1\)答案：AD3.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，且\(S_{n}=2a_{n}-2\)，則下列結(jié)論正確的是（）A.\(a_{2}=4\)B.數(shù)列\(zhòng)(\{a_{n}\}\)是等比數(shù)列C.\(a_{n}=2^{n}\)D.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=2^{n+1}-2\)答案：ABCD4.設(shè)\(\{a_{n}\}\)是等差數(shù)列，\(S_{n}\)為其前\(n\)項和，且\(S_{7}\ltS_{8}\)，\(S_{8}=S_{9}\gtS_{10}\)，則下列結(jié)論正確的是（）A.\(d\lt0\)B.\(a_{9}=0\)C.\(S_{11}\gtS_{7}\)D.\(S_{8}\)與\(S_{9}\)均為\(S_{n}\)的最大值答案：ABD5.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}=2a_{n}+1\)，則（）A.數(shù)列\(zhòng)(\{a_{n}+1\}\)是等比數(shù)列B.數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=2^{n}-1\)C.數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列D.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=2^{n+1}-n-2\)答案：ABC6.已知數(shù)列\(zhòng)(\{a_{n}\}\)是等比數(shù)列，公比\(q\gt1\)，前\(n\)項和為\(S_{n}\)，若\(a_{2}+a_{3}+a_{4}=28\)，\(a_{3}+2\)是\(a_{2}\)，\(a_{4}\)的等差中項，則下列說法正確的是（）A.\(q=2\)B.\(a_{1}=2\)C.\(S_{n}=2^{n+1}-2\)D.數(shù)列\(zhòng)(\{\log_{2}a_{n}\}\)的前\(n\)項和\(T_{n}=\frac{n(n+1)}{2}\)答案：AD7.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=\frac{a_{n}}{1+a_{n}}\)，\(a_{1}=1\)，則（）A.數(shù)列\(zhòng)(\{\frac{1}{a_{n}}\}\)是等差數(shù)列B.\(a_{n}=\frac{1}{n}\)C.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=\frac{n}{n+1}\)D.數(shù)列\(zhòng)(\{\frac{1}{a_{n}}\}\)的前\(n\)項和\(T_{n}=\frac{n(n+1)}{2}\)答案：ABD8.設(shè)\(\{a_{n}\}\)是首項為\(a_{1}\)，公差為\(d\)的等差數(shù)列，\(\{b_{n}\}\)是首項為\(b_{1}\)，公比為\(q\)的等比數(shù)列，則下列說法正確的是（）A.若\(a_{n}=b_{n}\)對\(n=1\)，\(2\)，\(3\)都成立，則\(a_{1}=b_{1}\)B.若\(a_{n}=b_{n}\)對\(n=1\)，\(2\)，\(3\)都成立，則\(d=b_{1}(q-1)\)C.若\(a_{1}=b_{1}\)，\(d=b_{1}(q-1)\)，則\(a_{n}=b_{n}\)對\(n\inN^{}\)都成立D.若\(a_{n}=b_{n}\)對\(n=1\)，\(2\)，\(3\)都成立，且\(q\neq1\)，則\(a_{1}=b_{1}\)，\(d=b_{1}(q-1)\)答案：ABD9.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}-6n\)，則（）A.\(\{a_{n}\}\)是等差數(shù)列B.\(a_{n}=2n-7\)C.\(S_{n}\)的最小值為\(-9\)D.使得\(S_{n}\lt0\)的\(n\)的最大值為\(5\)答案：ABC10.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}=a_{n}+n+1\)，則（）A.\(a_{n}=\frac{n(n+1)}{2}\)B.數(shù)列\(zhòng)(\{\frac{1}{a_{n}}\}\)的前\(n\)項和\(T_{n}=\frac{2n}{n+1}\)C.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=\frac{n(n+1)(n+2)}{6}\)D.數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列答案：ABD三、判斷題1.若數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=n^{2}\)，則\(a_{n}=2n-1\)。（√）2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}\lt0\)，\(q\gt1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列。（×）3.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(a_{6}=0\)，則\(S_{11}=0\)。（√）4.若數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=n\)，則\(a_{n}\)是等差數(shù)列。（×）5.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}+a_{2}=1\)，\(a_{3}+a_{4}=9\)，則公比\(q=3\)。（×）6.已知數(shù)列\(zhòng)(\{a_{n}\}\)的通項公式\(a_{n}=n^{2}-5n+4\)，則\(a_{n}\)的最小值為\(-\frac{9}{4}\)。（×）7.若數(shù)列\(zhòng)(\{a_{n}\}\)是等比數(shù)列，且\(a_{n}\gt0\)，\(a_{2}a_{4}+2a_{3}a_{5}+a_{4}a_{6}=25\)，則\(a_{3}+a_{5}=5\)。（√）8.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(S_{n}\)是其前\(n\)項和，若\(S_{10}=S_{20}\)，則\(S_{30}=0\)。（√）9.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}=2a_{n}\)，則\(a_{n}=2^{n-1}\)。（√）10.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{5}=2\)，\(a_{9}=8\)，則\(a_{7}=4\)。（√）四、簡答題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，\(a_{3}=7\)，\(S_{4}=24\)，求\(a_{n}\)與\(S_{n}\)。答案：設(shè)等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)。由\(a_{3}=7\)可得\(a_{1}+2d=7\)；由\(S_{4}=24\)可得\(4a_{1}+\frac{4\times3}{2}d=24\)，即\(4a_{1}+6d=24\)。聯(lián)立方程\(\begin{cases}a_{1}+2d=7\\4a_{1}+6d=24\end{cases}\)，解得\(\begin{cases}a_{1}=3\\d=2\end{cases}\)。所以\(a_{n}=a_{1}+(n-1)d=3+2(n-1)=2n+1\)，\(S_{n}=na_{1}+\frac{n(n-1)}{2}d=3n+n(n-1)=n^{2}+2n\)。2.已知等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=2\)