数列基础专项训练-2027届高三数学一轮复习

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高考一轮复习专项基础训练10--数列

本卷满分115分,时间75分钟

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一、单选题(共30分)

1.(本题5分)已知等差数列eqId83cf38189d5cbf627d2b82ac0eb76006满足eqId1dcc73a71b205259c160d0f1fd87ca86,eqId4e85b39c5434fefdc035ee0329f2d4e9,则eqId42e4487468ab2823d6dbf7f0ebd2eb38(   )

A.2 B.0 C.eqIdacbc6a613224461ade69362d46550474 D.eqId274a9dc37509f01c2606fb3086a46f4f

2.(本题5分)已知数列eqId83cf38189d5cbf627d2b82ac0eb76006的前eqIdb6a24198bd04c29321ae5dc5a28fe421项和公式为eqId944777f50704d7faf7a7d1fff013b4f1,则eqId42e4487468ab2823d6dbf7f0ebd2eb38(    )

A.eqIde8d02ea8c4988c5c28ab93f0d70fb55a B.eqIdd07ae0b4264da6a8812454ffd2f20d94 C.eqId2d0b71b8d2c183154221f717ce09077b D.eqId8da45c443af7994a26ffa9d8894e7262

3.(本题5分)已知eqId08eb71ecf8d733b6932f4680874dbbf3是等差数列eqId76aef4cdcb5af742ce28003b7b6c8c20的前n项和,若eqId8b1d3a06ee3a948a439f482794c8540c,则eqIdeb0abedd59476f8793d5a8948590a70c,求eqId5da4cd81500bdb43118150dbdb1541e6为(    )

A.4 B.5 C.6 D.7

4.(本题5分)在等比数列eqId63d471926f7b27322d90c82b9ce21d3d中,eqIdf966272f7781790ff27e40db6b525253,eqId4e85b39c5434fefdc035ee0329f2d4e9,则eqIddaf464629fa321a6ff7401ab79f07083等于(    )

A.eqId61128ab996360a038e6e64d82fcba004 B.eqId533a7b702ada1dd80123e4041271d521 C.eqId274a9dc37509f01c2606fb3086a46f4f D.eqId72b8bee40319fe80e512d221cfe252a7

5.(本题5分)已知正项等比数列eqId83cf38189d5cbf627d2b82ac0eb76006的前eqIdb6a24198bd04c29321ae5dc5a28fe421项和为eqId08eb71ecf8d733b6932f4680874dbbf3,且eqId00c3aed8080f03c6fbd5da85238278ab,eqId105fd92af4a217572cc971e485b450c4,则eqId614306cc3f34bdee4d5d885b79667645(    )

A.7 B.15 C.31 D.63

6.(本题5分)数列eqId76aef4cdcb5af742ce28003b7b6c8c20中,eqId302ae08810ef07e1ec320eb3ae8dbd48,eqId8fa6b54a8d26bc8eed83c8debfe61c49,则eqId047dbd9ff686703cad03aa383e5fec21(   ).

A.eqId4f71a7a540efafbbcedb18902536cf64 B.28 C.eqId8f88b597e0729b052743f9d8b7923b28 D.14

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二、多选题(共18分)

7.(本题6分)数列eqId83cf38189d5cbf627d2b82ac0eb76006为等差数列,eqId08eb71ecf8d733b6932f4680874dbbf3为其前eqIdb6a24198bd04c29321ae5dc5a28fe421项和.已知eqId996787ed4f0ce8d0d41948aa3ddecd6a,eqIdc95e18d1513022fffebfd84d052244e0,则下列结论正确的有(    )

A.公差eqIdae3012337aa392709349731fb1eef5b5 B.eqId42aab89e67736c81f63af477d8990dfd

C.eqId4198f60d3c8c7eb1497fe5b921d058b0 D.当eqId7c345907ebe27888332b1b44c666cc47时,eqId08eb71ecf8d733b6932f4680874dbbf3最小

8.(本题6分)记eqId08eb71ecf8d733b6932f4680874dbbf3为正项等比数列eqId63d471926f7b27322d90c82b9ce21d3d的前eqIdb6a24198bd04c29321ae5dc5a28fe421项和,eqId9aa8a716a31b0f51b70fdf9bdb257909为eqId63d471926f7b27322d90c82b9ce21d3d的公比,若eqId8af170367f036ba4d34743f07c1a998e,eqId4080bdff9b6a7c72f012e25172f2e889,则(    )

A.eqId2c8a5d993e275f2366c56861825f66a6 B.eqId1d59986e70f3b01a566d7522d0c0a7c1

C.eqIdcce224c28ca451c4f105dc3b077736cb是递增数列 D.eqIdb3882eeed3f29f68ab0df1ea98e6b025

9.(本题6分)已知eqId08eb71ecf8d733b6932f4680874dbbf3是等差数列eqId83cf38189d5cbf627d2b82ac0eb76006的前eqIdb6a24198bd04c29321ae5dc5a28fe421项和,且eqIdfd3c579e5e0540f190994cbb5b0653a6,eqId345eaa86e163b97a37552f636b587e73,则下列选项正确的是(    )

A.数列eqId83cf38189d5cbf627d2b82ac0eb76006为递减数列 B.eqIdff1508953a6b81848a89411596db8c00page number 0

 

试卷第2页(共2页)

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C.eqId08eb71ecf8d733b6932f4680874dbbf3的最大值为eqId194592cb77de8a597d5d64e1c85c3249 D.使得eqId49763402b2f2023f0ba64c37924267d3时eqIdb6a24198bd04c29321ae5dc5a28fe421的最大值是13

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三、填空题(共15分)

10.(本题5分)在等差数列eqId76aef4cdcb5af742ce28003b7b6c8c20中,eqIda83e47a80d6480847234a0047b43d193,则eqId7c98c59cd4749afdd21e73529fc84323______,eqId7999465d0e871febde66296a0cbf058c______.

11.(本题5分)已知数列eqId83cf38189d5cbf627d2b82ac0eb76006的前eqIdb6a24198bd04c29321ae5dc5a28fe421项和公式为eqId618374ba162083008c3b8ca296aadb6b,则eqId7999465d0e871febde66296a0cbf058c______;数列eqId83cf38189d5cbf627d2b82ac0eb76006的通项公式eqIde3f7fda69e2b32b9ced2239f915fa59b______.

12.(本题5分)已知数列eqId83cf38189d5cbf627d2b82ac0eb76006为等比数列,eqIdb065334d8f60c49f4bd3d9f1373fe4cd,eqId9cdfc4722a9dff9a8394b409ab9e6d7a,则eqId614306cc3f34bdee4d5d885b79667645__________.

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四、解答题(共52分)

13.(本题15分)设等差数列eqId83cf38189d5cbf627d2b82ac0eb76006的公差为eqId5c02bc0c74292b1e8f395f90935d3174,前eqIdb6a24198bd04c29321ae5dc5a28fe421项和为eqId08eb71ecf8d733b6932f4680874dbbf3,

(1)若eqIdfc807ea4b8a7ed325aee49aa292552ff,eqIdce018bfb9412c448d53c8e31b79efacc,求eqIde72adb45c60c2f63b46e65ff787302bf;

(2)若eqId5c75c05278f52012613d365dbd2faaf1,求eqId96abfe2da27a63e6affb19a0c80236d9;

(3)若eqId233fade8b59ccd59838a3f20d08f6b5a,求eqIdb6a24198bd04c29321ae5dc5a28fe421.

 

 

14.(本题17分)记数列eqId76aef4cdcb5af742ce28003b7b6c8c20的前n项和为eqId08eb71ecf8d733b6932f4680874dbbf3,已知eqId57c599e7cec6d192fb73218e7882ceca.

(1)求eqId96abfe2da27a63e6affb19a0c80236d9;

(2)记数列eqIdad83668ff336589f82a2cd04db9f9947的前n项和为eqIdf1ae9a3b0b7aeb1545b65d91aa371b3c,证明:eqIdc215db1d8f69757118ad405b78035628.

 

 

15.(本题20分)设数列eqId76aef4cdcb5af742ce28003b7b6c8c20的前n项和为eqId08eb71ecf8d733b6932f4680874dbbf3,已知eqIde3c07d6d0c63061e09e36b5a2c74760b.若eqId9b1959ed91572db2911072249ce8aa03.

(1)计算eqIde72adb45c60c2f63b46e65ff787302bf,eqId3e88093a749c0d46e0ee931ecfaff925,eqId6c1ccc6c74b8754e9bcbb3f39a11b6f1的值,并猜想数列eqId0f329b217e1051b23f0d61023cdc6e69的通项公式;

(2)证明:数列eqIdf5f75e87cdd8246a723b4f2b21351d81为等比数列,并求数列eqId76aef4cdcb5af742ce28003b7b6c8c20的通项公式;

(3)设eqIdef9864b8e099a5382cd351aa40ac7bfd,数列eqIdc5ab0309e2cd35585ea9fb2cc3017abf的前n项和为eqIdf1ae9a3b0b7aeb1545b65d91aa371b3c,求eqIdf1ae9a3b0b7aeb1545b65d91aa371b3c.

page number 1

 

试卷第2页(共1页)

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《高三一轮复习--数列专项基础训练(高考题型)》参考答案

题号

1

2

3

4

5

6

7

8

9

答案

A

B

B

A

C

B

BCD

BCD

AC

 

1.A

【详解】eqId16f3d198e76391779fa3badc848c8ac8数列eqId83cf38189d5cbf627d2b82ac0eb76006为等差数列,eqId1dcc73a71b205259c160d0f1fd87ca86,eqId4e85b39c5434fefdc035ee0329f2d4e9,

eqId42fdbeb90cb9a6768e230047343794f7,得eqIdc024a43158acfe36a43f12080ee581c0.

2.B

【详解】eqId3b368897a387c6b833d0b8c77db3c3a3.

3.B

【详解】对于等差数列eqIde5415c9d4eef2605dcd836f65a0ccd3b(eqId670d5fc71b49fdb5411b046bb9a81bab),其前eqIdb6a24198bd04c29321ae5dc5a28fe421项和满足公式eqId37db4da2b2da526c98f9bc428e6264d5,因此eqId113e2aa3ac7c162e7e5940ca8f82a035.

根据等差数列下标和性质:若eqIdb165cd9d2edb9104b73a5d7ac0ffbd35且eqId022ab8749fb27c2c66e750c40cc3e5e3,则eqId3fe3ce43ad186101b0f9758cc5163cd9,因此eqIda988fceaf43117dc124b722aa427f1f8.

将eqId00f74faf510ff069f56a8adfa1386bee、eqIda231cd5fb3c7ac4f08aa36069ba3d796代入前eqIdb6a24198bd04c29321ae5dc5a28fe421项和表达式:eqId60eee3c09e929b48ca4f0910ecd1a5d7,

代入eqIda231cd5fb3c7ac4f08aa36069ba3d796化简:eqId1aeea9aa7957ddd07fc58c1db67d80d6 ,解得eqId65a3858a32cc1e9151c7ea963b1a398b.

4.A

【详解】由等比数列性质可得:eqId483ba8c2af7526aefad05e7133f38e0c,即eqId9480c07502d33dd233dd87a1a6f8f4b8,则eqId3958dbf3c1defd700fd0ad0c636f3325.

5.C

【详解】设等比数列eqId76aef4cdcb5af742ce28003b7b6c8c20的公比为eqId2e420efb2afd06e85a595f837242ef61,且eqIde5a220a85883d0589001820e3a3c7c1e,

因为eqId0de851d172f030aae706c752930eb925,则eqId5c509b79e1e2fe58feb1042858ac269f,

又因为eqId00c3aed8080f03c6fbd5da85238278ab,则eqIdcf8c4060c7a614a2f287e9fa8fd9f627,可得eqId3518b5a336ac47321619900ae40b79d9,即eqIdf5f0b817512e1d4979c12c54db9fbc35,

所以eqId709b2954dcd4aa1953479fef9fb55e5e.

6.B

【详解】由eqId0b09840c2a408e9fb45d048983c452f9,eqId8fa6b54a8d26bc8eed83c8debfe61c49,

得eqIdfa8ce7f4da29c3ab454804c463e96fe8.

7.BCD

【分析】根据等差数列前eqIdb6a24198bd04c29321ae5dc5a28fe421项和公式,结合等差数列的通项公式逐一判断即可.

【详解】设等差数列eqId83cf38189d5cbf627d2b82ac0eb76006的公差为eqId5c02bc0c74292b1e8f395f90935d3174,则eqIdc9d551cb344a0976aa79369233df264b,故A错误;page number 2

答案第1页(共1页)

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