數(shù)列高考真題及答案

一、單項(xiàng)選擇題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和為\(S_{n}\)，若\(a_{3}=5\)，\(S_{9}=81\)，則\(a_{7}=(\)\)A.\(18\)B.\(13\)C.\(9\)D.\(7\)答案：B2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{5}=16\)，則公比\(q=(\)\)A.\(-2\)B.\(2\)C.\(\pm2\)D.\(4\)答案：B3.數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=2n-1\)，則\(a_{1}\)與\(a_{5}\)的等比中項(xiàng)為\((\)\)A.\(9\)B.\(\pm9\)C.\(3\)D.\(\pm3\)答案：D4.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}=a_{n}+2\)，則\(a_{10}=(\)\)A.\(17\)B.\(18\)C.\(19\)D.\(20\)答案：C5.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}+a_{5}=10\)，\(a_{4}=7\)，則數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\((\)\)A.\(1\)B.\(2\)C.\(3\)D.\(4\)答案：B6.等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和為\(S_{n}\)，若\(S_{3}=1\)，\(S_{6}=9\)，則\(S_{9}=(\)\)A.\(27\)B.\(28\)C.\(29\)D.\(30\)答案：B7.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}\)，則\(a_{8}=(\)\)A.\(15\)B.\(16\)C.\(49\)D.\(64\)答案：A8.在等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{3}+a_{4}+a_{5}+a_{6}+a_{7}=450\)，則\(a_{2}+a_{8}=(\)\)A.\(180\)B.\(270\)C.\(360\)D.\(540\)答案：A9.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{5}=4a_{3}\)，則\(a_{n}=(\)\)A.\(2^{n-1}\)B.\((-2)^{n-1}\)C.\(2^{n-1}\)或\((-2)^{n-1}\)D.\(2^{n}\)答案：C10.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=2a_{n}\)，且\(a_{1}=1\)，則\(a_{6}=(\)\)A.\(32\)B.\(16\)C.\(8\)D.\(64\)答案：A二、多項(xiàng)選擇題1.下列關(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\(a_{n}=kn+b\)（\(k,b\)為常數(shù)）B.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}\)是關(guān)于\(n\)的二次函數(shù)C.若\(\{a_{n}\}\)是等差數(shù)列，\(m,n,p,q\inN^+\)，且\(m+n=p+q\)，則\(a_{m}+a_{n}=a_{p}+a_{q}\)D.等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差\(d\gt0\)時(shí)，數(shù)列\(zhòng)(\{a_{n}\}\)是遞增數(shù)列答案：ACD2.設(shè)等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比為\(q\)，則下列說(shuō)法正確的是（）A.若\(q\gt1\)，則\(\{a_{n}\}\)是遞增數(shù)列B.若\(a_{1}\gt0\)，\(0\ltq\lt1\)，則\(\{a_{n}\}\)是遞減數(shù)列C.若\(a_{1}a_{9}=a_{5}^{2}\)D.等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)（\(q\neq1\)）答案：BCD3.已知數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列，\(S_{n}\)為其前\(n\)項(xiàng)和，若\(a_{3}=6\)，\(S_{5}=30\)，則（）A.首項(xiàng)\(a_{1}=2\)B.公差\(d=2\)C.\(a_{n}=2n\)D.\(S_{n}=n^{2}+n\)答案：ABCD4.對(duì)于數(shù)列\(zhòng)(\{a_{n}\}\)，若\(a_{n+1}-a_{n}=d\)（\(d\)為常數(shù)），則下列說(shuō)法正確的是（）A.數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列B.\(a_{n}=a_{1}+(n-1)d\)C.\(S_{n}=na_{1}+\frac{n(n-1)}{2}d\)D.若\(d\gt0\)，則\(S_{n}\)單調(diào)遞增答案：ABC5.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，已知\(a_{1}=1\)，\(a_{4}=8\)，則（）A.公比\(q=2\)B.\(a_{n}=2^{n-1}\)C.\(S_{n}=2^{n}-1\)D.\(a_{2}\cdota_{3}=8\)答案：ABCD6.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差\(d\neq0\)，\(a_{1}=1\)，且\(a_{1},a_{3},a_{9}\)成等比數(shù)列，則（）A.\(a_{n}=n\)B.\(S_{n}=\frac{n(n+1)}{2}\)C.數(shù)列\(zhòng)(\{\frac{1}{a_{n}a_{n+1}}\}\)的前\(n\)項(xiàng)和\(T_{n}=\frac{n}{n+1}\)D.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}\)答案：ABC7.下列數(shù)列中，是等比數(shù)列的有（）A.\(1,-1,1,-1,\cdots\)B.\(1,2,4,8,\cdots\)C.\(0,2,4,8,\cdots\)D.\(a,a,a,\cdots\)（\(a\neq0\)）答案：ABD8.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{5}=10\)，\(a_{12}=31\)，則（）A.\(a_{1}=-2\)B.\(d=3\)C.\(a_{n}=3n-5\)D.\(S_{n}=\frac{n(3n-7)}{2}\)答案：ABCD9.設(shè)等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和為\(S_{n}\)，公比\(q=2\)，\(S_{4}=15\)，則（）A.\(a_{1}=1\)B.\(a_{2}=2\)C.\(a_{3}=4\)D.\(S_{8}=255\)答案：ABCD10.已知數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=2n\)，\(a_{1}=1\)，則（）A.\(a_{n}=n^{2}-n+1\)B.\(a_{2}=3\)C.數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=\frac{n(n^{2}-n+3)}{3}\)D.\(a_{n+1}-a_{n}\)是等差數(shù)列答案：ABD三、判斷題1.若數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}-a_{n}=n\)，則\(\{a_{n}\}\)是等差數(shù)列。（×）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(xiàng)和\(S_{n}\)一定是關(guān)于\(n\)的二次函數(shù)。（×）4.若\(\{a_{n}\}\)是等比數(shù)列，\(a_{1}=1\)，\(a_{3}=4\)，則\(a_{5}=16\)。（√）5.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}+1\)，則\(a_{n}=2n-1\)。（×）6.等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比\(q\)可以為\(0\)。（×）7.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}+a_{7}=10\)，則\(a_{4}=5\)。（√）8.若數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=2a_{n}\)，則\(\{a_{n}\}\)是等比數(shù)列。（×）9.等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差\(d\)決定了數(shù)列的單調(diào)性。（√）10.等比數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}\)滿足\(S_{n},S_{2n}-S_{n},S_{3n}-S_{2n}\)仍成等比數(shù)列。（×）四、簡(jiǎn)答題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{7}=13\)，求數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}\)。答案：首先求公差\(d\)，\(d=\frac{a_{7}-a_{3}}{7-3}=\frac{13-5}{4}=2\)。然后由\(a_{n}=a_{m}+(n-m)d\)，這里\(m=3\)，則\(a_{n}=a_{3}+(n-3)d=5+2(n-3)=2n-1\)。所以數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式為\(a_{n}=2n-1\)。2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{4}=16\)，求公比\(q\)和前\(n\)項(xiàng)和\(S_{n}\)。答案：由等比數(shù)列通項(xiàng)公式\(a_{n}=a_{1}q^{n-1}\)，已知\(a_{1}=2\)，\(a_{4}=16\)，則\(a_{4}=a_{1}q^{3}\)，即\(16=2q^{3}\)，解得\(q=2\)。等比數(shù)列前\(n\)項(xiàng)和公式\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)（\(q\neq1\)），把\(a_{1}=2\)，\(q=2\)代入得\(S_{n}=\frac{2(1-2^{n})}{1-2}=2^{n+1}-2\)。3.已知數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}-2n\)，求數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式。答案：當(dāng)\(n=1\)時(shí)，\(a_{1}=S_{1}=1^{2}-2\times1=-1\)；當(dāng)\(n\geq2\)時(shí)，\(a_{n}=S_{n}-S_{n-1}=n^{2}-2n-[(n-1)^{2}-2(n-1)]\)，化簡(jiǎn)得\(a_{n}=n^{2}-2n-(n^{2}-2n+1-2n+2)=2n-3\)。當(dāng)\(n=1\)時(shí)，\(2\times1-3=-1=a_{1}\)，所以\(a_{n}=2n-3\)。4.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和為\(S_{n}\)，已知\(a_{1}=1\)，\(S_{3}=9\)，求\(S_{n}\)的表達(dá)式。答案：設(shè)等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)，由\(S_{n}=na_{1}+\frac{n(n-1)}{2}d\)，且\(S_{3}=9\)，\(a_{1}=1\)，可得\(S_{3}=3\times1+\frac{3\times(3-1)}{2}d=9\)，即\(3+3d=9\)，解得\(d=2\)。將\(a_{1}=1\)，\(d=2\)代入\(S_{n}\)公式，得\(S_{n}=n\times1+\frac{n(n-1)}{2}\times2=n^{2}\)。五、討論題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)和等比數(shù)列\(zhòng)(\{b_{n}\}\)，\(a_{1}=b_{1}=1\)，\(a_{2}=b_{2}\)，\(a_{4}=b_{3}\)，討論數(shù)列\(zhòng)(\{a_{n}\}\)和\(\{b_{n}\}\)的通項(xiàng)公式