吉林高考數(shù)列題真題及答案

一、單項(xiàng)選擇題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{5}=9\)，則\(a_{7}\)的值為（）A.11B.12C.13D.14答案：C2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=2\)，\(a_{5}=16\)，則公比\(q\)為（）A.-2B.2C.\(\pm2\)D.4答案：B3.數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=2n-1\)，則\(a_{5}\)的值是（）A.9B.10C.11D.12答案：A4.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和為\(S_{n}\)，若\(a_{1}=1\)，\(S_{3}=9\)，則公差\(d\)等于（）A.1B.2C.3D.4答案：B5.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{4}=8\)，則\(a_{3}\)的值為（）A.4B.\(\pm4\)C.2D.\(\pm2\)答案：A6.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}=a_{n}+2\)，則\(a_{10}\)的值為（）A.17B.18C.19D.20答案：C7.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}+a_{4}=10\)，則\(a_{3}\)的值為（）A.5B.6C.7D.8答案：A8.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}a_{5}=16\)，則\(a_{4}\)的值為（）A.4B.\(\pm4\)C.8D.\(\pm8\)答案：B9.數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=n^{2}-n\)，則\(a_{3}\)的值是（）A.3B.6C.9D.12答案：B10.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和為\(S_{n}\)，若\(S_{5}=25\)，則\(a_{3}\)的值為（）A.5B.6C.7D.8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\(a_{n+1}-a_{n}\)是常數(shù)B.等差數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=a_{1}+(n-1)d\)C.若\(a,b,c\)成等差數(shù)列，則\(2b=a+c\)D.等差數(shù)列的前\(n\)項(xiàng)和公式\(S_{n}=\frac{n(a_{1}+a_{n})}{2}\)答案：ABCD2.等比數(shù)列\(zhòng)(\{a_{n}\}\)滿足（）A.\(\frac{a_{n+1}}{a_{n}}=q\)（\(q\)為常數(shù)，\(q\neq0\)）B.\(a_{n}=a_{1}q^{n-1}\)C.若\(m,n,p,q\inN^+\)，\(m+n=p+q\)，則\(a_{m}a_{n}=a_{p}a_{q}\)D.等比數(shù)列的前\(n\)項(xiàng)和公式\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)（\(q\neq1\)）答案：ABCD3.已知數(shù)列\(zhòng)(\{a_{n}\}\)是等差數(shù)列，\(a_{3}=5\)，\(a_{7}=13\)，則（）A.公差\(d=2\)B.\(a_{1}=1\)C.\(a_{5}=9\)D.前\(n\)項(xiàng)和\(S_{n}=n^{2}\)答案：ABC4.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(q=3\)，則（）A.\(a_{n}=2\times3^{n-1}\)B.\(a_{3}=18\)C.前\(n\)項(xiàng)和\(S_{n}=3^{n}-1\)D.\(a_{2}a_{4}=a_{3}^{2}\)答案：ABCD5.對(duì)于數(shù)列\(zhòng)(\{a_{n}\}\)，下列說(shuō)法正確的是（）A.若\(a_{n+1}-a_{n}=n\)，則\(\{a_{n}\}\)不是等差數(shù)列B.若\(a_{n}=3^{n}\)，則\(\{a_{n}\}\)是等比數(shù)列C.若\(a_{n}=n^{2}\)，則\(\{a_{n}\}\)既不是等差數(shù)列也不是等比數(shù)列D.若\(\{a_{n}\}\)是等差數(shù)列，\(a_{1}=1\)，\(a_{3}=5\)，則\(a_{n}=2n-1\)答案：ABCD6.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和為\(S_{n}\)，若\(a_{1}=-1\)，\(S_{3}=3\)，則（）A.公差\(d=2\)B.\(a_{2}=1\)C.\(S_{5}=15\)D.\(a_{n}=2n-3\)答案：ABCD7.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{2}=4\)，\(a_{5}=32\)，則（）A.公比\(q=2\)B.\(a_{1}=2\)C.\(a_{4}=16\)D.前\(n\)項(xiàng)和\(S_{n}=2^{n+1}-2\)答案：ABCD8.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{1}=1\)，\(a_{n+1}=2a_{n}+1\)，則（）A.\(a_{2}=3\)B.\(a_{3}=7\)C.數(shù)列\(zhòng)(\{a_{n}+1\}\)是等比數(shù)列D.\(a_{n}=2^{n}-1\)答案：ABCD9.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)，\(a_{n}=2n-5\)，則（）A.\(a_{1}=-3\)B.公差\(d=2\)C.\(S_{n}=n^{2}-4n\)D.\(a_{3}=1\)答案：ABCD10.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{3}=4\)，則（）A.公比\(q=2\)或\(q=-2\)B.\(a_{2}=2\)或\(a_{2}=-2\)C.\(a_{4}=8\)或\(a_{4}=-8\)D.前\(n\)項(xiàng)和\(S_{n}=\frac{1-2^{n}}{1-2}\)（當(dāng)\(q=2\)時(shí)）答案：ABC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，\(a_{1}=1\)，\(a_{2}=2\)，則公比\(q=2\)。（√）3.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=n^{2}+1\)，則\(\{a_{n}\}\)不是等差數(shù)列。（√）4.若\(a,b,c\)成等比數(shù)列，則\(b^{2}=ac\)。（√）5.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。（√）6.等差數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{m}+a_{n}=a_{p}+a_{q}\)，則\(m+n=p+q\)。（×）7.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{n}=a_{m}q^{n-m}\)。（√）8.數(shù)列\(zhòng)(\{a_{n}\}\)滿足\(a_{n+1}=a_{n}+2\)，\(a_{1}=1\)，則\(a_{n}=2n-1\)。（√）9.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}\)，若\(S_{n}=An^{2}+Bn\)（\(A,B\)為常數(shù)），則\(\{a_{n}\}\)是等差數(shù)列。（√）10.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(a_{2}=-1\)，\(a_{3}=1\)，則公比\(q=-1\)。（√）四、簡(jiǎn)答題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=3\)，\(a_{5}=11\)，求該數(shù)列的通項(xiàng)公式\(a_{n}\)。答案：首先求公差\(d\)，根據(jù)等差數(shù)列通項(xiàng)公式\(a_{n}=a_{1}+(n-1)d\)，\(a_{5}=a_{1}+4d\)。已知\(a_{1}=3\)，\(a_{5}=11\)，則\(11=3+4d\)，解得\(d=2\)。那么通項(xiàng)公式\(a_{n}=a_{1}+(n-1)d=3+2(n-1)=2n+1\)。2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=2\)，\(a_{3}=8\)，求公比\(q\)及\(a_{n}\)。答案：由等比數(shù)列通項(xiàng)公式\(a_{n}=a_{1}q^{n-1}\)可得\(a_{3}=a_{1}q^{2}\)。已知\(a_{1}=2\)，\(a_{3}=8\)，則\(8=2q^{2}\)，解得\(q^{2}=4\)，所以\(q=\pm2\)。當(dāng)\(q=2\)時(shí)，\(a_{n}=2\times2^{n-1}=2^{n}\)；當(dāng)\(q=-2\)時(shí)，\(a_{n}=2\times(-2)^{n-1}\)。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)，\(a_{n}=S_{n}-S_{n-1}=n^{2}-(n-1)^{2}=n^{2}-(n^{2}-2n+1)=2n-1\)。當(dāng)\(n=1\)時(shí)，\(2\times1-1=1=a_{1}\)，所以\(a_{n}=2n-1\)，\(n\inN^+\)。4.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和為\(S_{n}\)，已知\(a_{3}=5\)，\(S_{5}=25\)，求\(a_{n}\)。答案：由\(S_{5}=\frac{5(a_{1}+a_{5})}{2}=25\)，可得\(a_{1}+a_{5}=10\)。又因?yàn)樵诘炔顢?shù)列中\(zhòng)(a_{1}+a_{5}=2a_{3}\)，已知\(a_{3}=5\)，所以符合。設(shè)公差為\(d\)，\(a_{3}=a_{1}+2d=5\)。\(S_{5}=5a_{1}+\frac{5\times4}{2}d=25\)，即\(5a_{1}+10d=25\)，化簡(jiǎn)得\(a_{1}+2d=5\)，解得\(a_{1}=1\)，\(d=2\)，則\(a_{n}=1+2(n-1)=2n-1\)。五、討論題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，并舉例說(shuō)明。答案：等差數(shù)列在實(shí)際生活中應(yīng)用廣泛，比如在計(jì)算均勻增加或減少的數(shù)量問(wèn)題上。例如，一個(gè)劇場(chǎng)設(shè)置了\(20\)排座位，第一排有\(zhòng)(38\)個(gè)座位，往后每一排都比前一排多\(2\)個(gè)座位，就可以用等差數(shù)列來(lái)計(jì)算各排座位數(shù)及總座位數(shù)。等比數(shù)列常用于涉及增長(zhǎng)率或衰減率的情況，像銀行復(fù)利問(wèn)題，本金為\(P\)，年利率為\(r\)，存\(n\)年，本利和就是以\(P\)為首項(xiàng)，\((1+r)\)為公比的等比數(shù)列的第\(n+1\)項(xiàng)。2.已知數(shù)列\(zhòng)(\{a_{n}\}\)的通項(xiàng)公式\(a_{n}=n^{2}-5n+4\)，討論數(shù)列\(zhòng)(\{a_{n}\}\)的單調(diào)性。答案：要討論數(shù)列單調(diào)性，可通過(guò)比較\(a_{n+1}\)與\(a_{n}\)的大小。\(a_{n+1}-a_{n}=[(n+1)^{2}-5(n+1)+4]-(n^{2}-5n+4)\)，展開(kāi)得\(a_{n+1}-a_{n}=n^{2}+2n+1-5n-5+4-n^{2}+5n-4=2