题目

Problem

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.

In an arithmetic series,

the sixth term is $2$
the sum of the first ten terms is $-80$

For this series,

(a) find the value of the first term and the value of the common difference.

(4)

(b) Hence find the smallest value of $n$ for which

S_n>8000

(3)

解答

(a)

解法一

思路

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设首项为 $a$ ，公差为 $d$ 。第六项给一个方程，前十项和再给一个方程，联立即可。

答题过程

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The sixth term is $2$ , so

\begin{align*} a+5d=2. \end{align*}

Also,

\begin{align*} S_{10}=\frac{10}{2}(2a+9d)=-80. \end{align*}

Thus

\begin{align*} 5(2a+9d)=&\,-80\\ 2a+9d=&\,-16. \end{align*}

Solve

\begin{align*} a+5d=&\,2,\\ 2a+9d=&\,-16. \end{align*}

Doubling the first equation gives

\begin{align*} 2a+10d=4. \end{align*}

Subtract:

\begin{align*} d=&\,20,\\ a+5(20)=&\,2,\\ a=&\,-98. \end{align*}

Therefore,

\begin{align*} a=-98,\qquad d=20. \end{align*}

(b)

解法一

思路

展开

把 $a=-98$ ， $d=20$ 代入 $S_n$ 公式，解不等式。最后要取最小整数 $n$ 。

答题过程

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Using

\begin{align*} S_n=\frac n2(2a+(n-1)d), \end{align*}

we need

\begin{align*} \frac n2(2(-98)+(n-1)20) >&\,\, 8000\\[4mm] \frac n2(-196+20n-20) >&\,\, 8000\\[4mm] \frac n2(20n-216) >&\,\, 8000\\[4mm] n(10n-108) >&\,\, 8000\\[4mm] 10n^2-108n-8000 >&\,\, 0\\[4mm] 5n^2-54n-4000 >&\,\, 0. \end{align*}

Solve

\begin{align*} 5n^2-54n-4000=0. \end{align*}

The positive root is $n \approx 34.2$ .

Therefore, the smallest integer value of $n$ is $35$ .

解法二

思路

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试错法（Trial and improvement）。由于我们只需要求出满足 $S_n > 8000$ 的最小整数 $n$ ，代入 $a=-98$ 和 $d=20$ 得到 $S_n = 10n^2 - 108n$ 。利用计算器直接测试邻近的整数值，找到分界点即可。

答题过程

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From part (a), we have $a = -98$ and $d = 20$ .

The sum of the first $n$ terms is given by:

\begin{align*} S_n =&\,\, \frac{n}{2}[2(-98) + (n-1)20]\\[4mm] =&\,\, n[-98 + 10(n-1)]\\[4mm] =&\,\, n(10n - 108)\\[4mm] =&\,\, 10n^2 - 108n \end{align*}

We want to find the smallest integer $n$ such that $S_n > 8000$ .

Let’s test integer values for $n$ using a calculator:

For $n = 34$ :

\begin{align*} S_{34} =&\,\, 10(34)^2 - 108(34)\\[4mm] =&\,\, 11560 - 3672\\[4mm] =&\,\, 7888 < 8000 \end{align*}

For $n = 35$ :

\begin{align*} S_{35} =&\,\, 10(35)^2 - 108(35)\\[4mm] =&\,\, 12250 - 3780\\[4mm] =&\,\, 8470 > 8000 \end{align*}

Since $S_{34} < 8000$ and $S_{35} > 8000$ , the smallest integer value of $n$ for which the sum exceeds $8000$ is $35$ .

IAL 2024 May Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程