题目

Problem

In a large theatre there are $n$ rows of seats, where $n$ is a constant.

The number of seats in the first row is $a$ , where $a$ is a constant.

In each subsequent row there are 4 more seats than in the previous row so that

Given that the total number of seats in the first 10 rows is 360,

(a) find the value of $a$ .

Given also that the total number of seats in the $n$ rows is 2146,

(b) show that $n^2 + 8n - 1073 = 0$ .

(i) state the number of rows of seats in the theatre,

(ii) find the maximum number of seats in any one row.

(7)

题目中文翻译

一个大剧院有 $n$ 排座位，其中 $n$ 是常数。

第一排有 $a$ 个座位，其中 $a$ 是常数。

每一排比前一排多4个座位，因此：

已知前10排的总座位数为360，

(a) 求 $a$ 的值。

又已知 $n$ 排的总座位数为2146，

(b) 证明 $n^2 + 8n - 1073 = 0$ 。

(i) 求剧院的排数，

(ii) 求任一排的最大座位数。

（7分）

解答

等差数列求和公式 $S_n = \dfrac{n}{2}[2a + (n-1)d]$ ，其中 $a$ 为首项， $d = 4$ 为公差。

(a) The total number of seats in the first 10 rows is 360. Using the sum formula with $d = 4$ :

S_{10} =&\, \frac{10}{2}[2a + 9 \times 4] \\ 360 =&\, 5[2a + 36] \\ 72 =&\, 2a + 36 \\ 2a =&\, 36 \\ a =&\, 18 \end{align*}$$ $$\boxed{a = 18}$$ **(b)** The total number of seats in $n$ rows is 2146: $$\begin{align*} S_n =&\, \frac{n}{2}[2 \times 18 + (n-1) \times 4] \\ 2146 =&\, \frac{n}{2}[36 + 4n - 4] \\ 2146 =&\, \frac{n}{2}[4n + 32] \\ 2146 =&\, n(2n + 16) \\ 2146 =&\, 2n^2 + 16n \\ 2n^2 + 16n - 2146 =&\, 0 \\ n^2 + 8n - 1073 =&\, 0 \end{align*}$$ $$\boxed{n^2 + 8n - 1073 = 0}$$ **(c)(i)** Solving $n^2 + 8n - 1073 = 0$: $$\begin{align*} n =&\, \frac{-8 \pm \sqrt{64 + 4292}}{2} = \frac{-8 \pm 66}{2} \end{align*}$$ Taking the positive value: $n = \dfrac{58}{2} = 29$ $$\boxed{n = 29}$$ **(c)(ii)** The number of seats in the $n$th row: $$\begin{align*} a_{29} =&\, 18 + (29 - 1) \times 4 \\ =&\, 18 + 112 \\ =&\, 130 \end{align*}$$ $$\boxed{130 \text{ seats}}$$