《2025年优化探究同步导学案高中数学选择性必修第二册人教版》

答案： $\frac{n(a_1 + a_n)}{2}$，$na_1 + \frac{n(n - 1)}{2}d$。

例1
在公差为d的等差数列$\{ a_{n}\}$中.
(1)$a_{1}=4,S_{8}=172$,求$a_{8}$和d;
(2)已知$d = 2,a_{n}=11,S_{n}=35$,求$a_{1}$和n;
(3)已知等差数列$\{ a_{n}\}$的前n项和为$S_{n},S_{4}=40,S_{n}=210,S_{n - 4}=130$,求n.

答案： [例1][解]
(1)由已知，得$S_8 = \frac{8(a_1 + a_8)}{2}$
$ = \frac{8(4 + a_8)}{2} = 172$，解得$a_8 = 39$.
又$\because a_8 = 4 + (8 - 1)d = 39$，$\therefore d = 5$.
(2)由$\begin{cases}a_n = a_1 + (n - 1)d, \\S_n = na_1 + \frac{n(n - 1)}{2}d,\end{cases}$
得$\begin{cases}a_1 + 2(n - 1) = 11, \\na_1 + \frac{n(n - 1)}{2} × 2 = 35,\end{cases}$
解方程组得$\begin{cases}n = 5, \\a_1 = 3\end{cases}$或$\begin{cases}n = 7, \\a_1 = -1.\end{cases}$
(3)$S_n - S_{n - 4} = a_{n - 3} + a_{n - 2} + a_{n - 1} + a_n = 80$，$S_4 = a_1 + a_2 + a_3 + a_4 = 40$.
两式相加得$4(a_1 + a_n) = 120$，
$\therefore a_1 + a_n = 30$.
又$\because S_n = \frac{n(a_1 + a_n)}{2} = 15n = 210$，$\therefore n = 14$.

跟踪训练
1. 在等差数列$\{ a_{n}\}$中.
(1)已知$a_{6}=10,S_{5}=5$,求$a_{8}$和$S_{10}$;
(2)$a_{3}+a_{15}=40$,求$S_{17}$;
(3)$a_{1}=\frac{5}{6},a_{n}=-\frac{3}{2},S_{n}=-5$,求n和d.

答案： 跟踪训练1.解：
(1)$S_5 = 5a_1 + \frac{5 × 4}{2}d = 5$，解
得$\begin{cases}a_1 = -5, \\d = 3.\end{cases}$
$\therefore a_8 = a_1 + 2d = 10 + 2 × 3 = 16$，$S_{10} = 10a_1 + \frac{10 × 9}{2}d = 10 × (-5) + 5 × 9 × 3 = 85$.
(2)$S_{17} = \frac{17 × (a_1 + a_{17})}{2} = \frac{17 × (a_3 + a_{15})}{2}$
$ = \frac{17 × 40}{2} = 340$.
(3)由题意得，$S_n = \frac{n(a_1 + a_n)}{2} =$
$n(\frac{5}{6} - \frac{3}{2}) = -5$，解得$n = 15$.
又因为$a_{15} = \frac{5}{6} + (15 - 1)d = -\frac{3}{2}$，解得$d$
$ = -\frac{1}{6}$，所以$n = 15$，$d = -\frac{1}{6}$.