《2025年练习生高中数学选择性必修第二册人教版》

12. 已知等差数列$\{ a_n \}$中,$a_5 = 4$,公差$d = 4$. 若在$\{ a_n \}$中每相邻两项之间都插入两个数,使它们和原数列的数一起构成一个新的等差数列,求新数列的第 50 项及$a_{50}$是新数列的第几项?
题型 4 等差数列的性质

答案： 12.【解】由题意,得$a_{n} = a_{4} + (n - 5)d = 4n - 16$.
设新的等差数列为$\{ b_{n}\}$,其公差为$d'$,
则$b_{1} = a_{1} = - 12$,$d' = \frac {1}{3}d = \frac {4}{3}$,
$\therefore b_{n} = - 12 + \frac {4}{3}(n - 1) = \frac {4}{3}n - \frac {40}{3}$,$\therefore b_{50} = \frac {160}{3}$.
由$a_{50} = 184 = \frac {4}{3}n - \frac {40}{3}$,得$n = 148$.
$\therefore a_{50}$是新数列的第148项.

13. [2022·江西九江高二期末]在等差数列$\{ a_n \}$中,$a_3 + a_9 = 12$,$a_2 = 4$,则$a_{10} =$ (B)

A.4
B.8
C.3
D.6

答案： 13.B 【解析】方法一 因为$\{ a_{n}\}$为等差数列,所以$a_{5} + a_{9} = a_{2} + a_{10}$(若$m + n = p + q$,$m,n,p,q \in N^{*}$,则$a_{m} + a_{n} = a_{p} + a_{q}$),即$12 = 4 + a_{10}$,解得$a_{10} = 8$. 故选B.
方法二 因为$\{ a_{n}\}$为等差数列,所以$a_{3} + a_{9} = 2a_{6}$(等差中项)$= 12$,所以$a_{6} = 6$. 易得$a_{2},a_{6},a_{10}$构成等差数列(下标成等差数列的项构成等差数列),即$4,6,a_{10}$构成等差数列,所以$a_{10} = 8$. 故选B.
方法三 设等差数列$\{ a_{n}\}$的公差为$d$,则$\begin{cases}(a_{1} + 2d) + (a_{1} + 8d) = 12, \\a_{1} + d = 4,\end{cases}$解得$\begin{cases}a_{1} = \frac {7}{2}, \\d = \frac {1}{2}.\end{cases}$故$a_{10} = a_{1} + 9d = \frac {7}{2} + 9 × \frac {1}{2} = 8$. 选B.

14. [2022·江苏省仪征中学高二期末]设数列$\{ a_n \}$,$\{ b_n \}$都是等差数列,且$a_1 = 25$,$b_1 = 75$,$a_2 + b_2 = 100$,则$a_{2022} + b_{2022} =$ (C)

A.-2 022
B.0
C.100
D.2 022

答案： 14.C 【解析】因为数列$\{ a_{n}\}$,$\{ b_{n}\}$都是等差数列,所以数列$\{ a_{n} + b_{n}\}$也是等差数列. 又$a_{1} + b_{1} = 25 + 75 = 100$,$a_{2} + b_{2} = 100$,所以数列$\{ a_{n} + b_{n}\}$的公差为0,首项为100,所以$a_{n} + b_{n} = 100$,所以$a_{2022} + b_{2022} = 100$. 故选C.

15. 已知数列$\{ a_n \}$是等差数列,若$a_4 + a_7 + a_{10} = 17$,$a_4 + a_5 + a_6 + ·s + a_{13} + a_{14} = 77$,则公差$d =$ (D)

A.1
B.$\frac {1}{2}$
C.$\frac {1}{3}$
D.$\frac {2}{3}$

答案： 15.D 【解析】因为$a_{4} + a_{7} + a_{10} = 3a_{7} = 17$,所以$a_{7} = \frac {17}{3}$. 因为$a_{4} + a_{5} + a_{6} + ·s + a_{13} + a_{14} = 11a_{9} = 77$,所以$a_{9} = 7$,所以公差$d = \frac {a_{9} - a_{7}}{9 - 7} = \frac {2}{3}$. 故选D.
方法总结 在等差数列$\{ a_{n}\}$中,首项为$a_{1}$,公差为$d$,则$a_{n} = a_{1} + (n - 1)d$,且对任意的$m,n \in N^{*}$,有$a_{m} = a_{n} + (m - n)d$. 特别地,当$m \neq n$时,$d = \frac {a_{m} - a_{n}}{m - n}$,这表明已知等差数列中的任意两项即可求得其公差.

16. 已知等差数列$\{ a_n \}$单调递增且满足$a_1 + a_8 = 6$,则$a_6$的取值范围是 (C)

A.$(- \infty,3)$
B.$(3,6)$
C.$(3,+ \infty)$
D.$(6,+ \infty)$

答案： 16.C 【解析】设等差数列$\{ a_{n}\}$的公差为$d$. 因为数列$\{ a_{n}\}$单调递增,所以$d ＞ 0$,所以$a_{1} + a_{8} = a_{3} + a_{6} = 2a_{6} - 3d = 6$,则$2a_{6} - 6 = 3d ＞ 0$,解得$a_{6} ＞ 3$. 故选C.

17. [2023·甘肃天水麦积高二期末]已知等差数列$\{ a_n \}$中,$a_4 + a_7 + a_{10} = 18$,$a_6 + a_8 + a_{10} = 27$. 若$a_k = 21$,则$k =$12.

答案： 17.12 【解析】因为$a_{4} + a_{7} + a_{10} = 18$,且$a_{4} + a_{10} = 2a_{7}$,所以$a_{7} = 6$. 因为$a_{6} + a_{8} + a_{10} = 27$,$a_{6} + a_{10} = 2a_{8}$,所以$a_{8} = 9$. 所以公差$d = a_{8} - a_{7} = 3$,所以$a_{k} = a_{7} + (k - 7)d = 21$,即$6 + 3(k - 7) = 21$,解得$k = 12$.

18. [2023·河南省实验中学高二期中]在数列$\{ a_n \}$中,$a_1 = 1$,$a_{n + 1} = \frac {2a_n}{a_n + 2}$($n$为正整数),则$a_n =$ (B)

A.$\frac {n + 1}{2}$
B.$\frac {2}{n + 1}$
C.$\frac {2n}{n + 1}$
D.$\frac {n + 1}{2n}$

答案： 18.B 【解析】因为$a_{n + 1} = \frac {2a_{n}}{a_{n} + 2}$,所以$\frac {1}{a_{n + 1}} = \frac {a_{n} + 2}{2a_{n}} = \frac {1}{2} + \frac {1}{a_{n}}$. 因为$a_{1} = 1$,所以$\frac {1}{a_{1}} = 1$,所以数列$\{ \frac {1}{a_{n}}\}$是首项为1,公差为$\frac {1}{2}$的等差数列,所以$\frac {1}{a_{n}} = 1 + \frac {1}{2}(n - 1) = \frac {n + 1}{2}$,所以$a_{n} = \frac {2}{n + 1}$. 故选B.

19. 已知数列$\{ a_n \}$中,$a_1 = 1$,$\frac {1}{a_{n + 1}} = 1 + \frac {1}{a_n}$($n\in N^*$),则$a_{10} =$ (D)

A.$\frac {1}{7}$
B.$\frac {1}{8}$
C.$\frac {1}{9}$
D.$\frac {1}{10}$

答案： 19.D 【解析】令$b_{n} = \frac {1}{a_{n}}(n \in N^{*})$. 因为$a_{1} = 1$,$\frac {1}{a_{n + 1}} = 1 + \frac {1}{a_{n}}(n \in N^{*})$,所以$b_{n + 1} = 1 + b_{n}$,$b_{1} = 1$,所以数列$\{ b_{n}\}$是以1为首项,1为公差的等差数列,所以$b_{n} = n$,则$a_{n} = \frac {1}{n}$,所以$a_{10} = \frac {1}{10}$. 故选D.

20. [2023·湖南常德临澧第一中学高二入学考试]已知数列$\{ a_n \}$的首项$a_1 = 0$,$a_{n + 1} = a_n + 2\sqrt {a_n + 1} + 1$,则$a_{20} =$ (A)

A.399
B.401
C.404
D.901

答案： 20.A 【解析】由题意,知$a_{n + 1} + 1 = (\sqrt {a_{n} + 1} + 1)^{2}$,则$\sqrt {a_{n + 1} + 1} - \sqrt {a_{n} + 1} = 1$,所以数列$\{ \sqrt {a_{n} + 1}\}$是以1为首项,1为公差的等差数列,所以$\sqrt {a_{n} + 1} = n$,则$a_{n} = n^{2} - 1$,所以$a_{20} = 20^{2} - 1 = 399$. 故选A.

21. 已知数列$\{ a_n \}$的各项均为正数,$a_1 = 2$,$a_{n + 1} - a_n = \frac {4}{a_{n + 1} + a_n}$,则$a_n =$$2\sqrt{n}$.

答案： 21.$2\sqrt{n}$ 【解析】由题意,得$a_{n + 1}^{2} - a_{n}^{2} = 4$,$a_{1}^{2} = 4$,所以数列$\{ a_{n}^{2}\}$是以4为首项,4为公差的等差数列,所以$a_{n}^{2} = 4 + 4(n - 1) = 4n$,则$a_{n} = 2\sqrt{n}$.

22. 四个数成等差数列,其平方和为 94,第一个数与第四个数的积比第二个数与第三个数的积少 18,求这四个数构成的数列.

答案： 22.【解】设四个数构成的数列为$a - 3d$,$a - d$,$a + d$,$a + 3d$.根据题意,得$(a - 3d)^{2} + (a - d)^{2} + (a + d)^{2} + (a + 3d)^{2} = 94$,
则$2a^{2} + 10d^{2} = 47$.①
又$(a - 3d)(a + 3d) = (a - d)(a + d) - 18$,
整理得$8d^{2} = 18$,解得$d = \pm \frac{3}{2}$.代入①得$a = \pm \frac{7}{2}$,
故所求四个数构成的数列为8,5,2,−1或1,−2,−5,−8 或−1,2,5,8或−8,−5,−2,1.
易错规避
(1)若四个数成等差数列,可设为$a - 3d$,$a - d$,$a + d$,$a + 3d$,此时易误认为数列的公差为$d$;
(2)作答时易漏解,原因在于没有按照数列的形式表示所求的四个数.