题目

Problem

Jem pays money into a savings scheme, A, over a period of $300$ months. Jem pays £20 into scheme A in month 1, £20.50 in month 2, £21 in month 3 and so on, so that the amounts Jem pays each month form an arithmetic sequence.

(a) Show that Jem pays £69.50 into scheme A in month 100

(1)

(b) Find the total amount that Jem pays into scheme A over the period of $300$ months.

(2)

Kim pays money into a different savings scheme, B, over the same period of $300$ months. In a model, the amounts Kim pays into scheme B increase by the same percentage each month, so that the amounts Kim pays each month form a geometric sequence. Given that Kim pays

£20 into scheme B in month 1
£250 into scheme B in month 300

(c) use the model to calculate, to the nearest £10, the difference between the total amount paid into scheme A and the total amount paid into scheme B over the period of $300$ months.

(3)

题目中文翻译

Jem 在 $300$ 个月内向储蓄计划 A 存款。第1个月存入 £20，第2个月 £20.50，第3个月 £21，依此类推，每月存款金额构成等差数列。

(a) 证明 Jem 在第100个月向计划 A 存入 £69.50。(1)

(b) 求 Jem 在这 $300$ 个月内向计划 A 存入的总金额。(2)

Kim 在同样的 $300$ 个月内向另一储蓄计划 B 存款。在一个模型中，Kim 每月的存款金额按相同百分比增长，因而构成等比数列。已知 Kim：

第1个月向计划 B 存入 £20；
第300个月向计划 B 存入 £250。

解答

(a)

解法一

思路

展开

scheme A 是等差数列，首项 $20$ ，公差 $0.5$ 。

答题过程

展开

For scheme A,

\begin{align*} u_{100}=&\,20+(100-1)(0.5)\\ =&\,20+49.5\\ =&\,69.50. \end{align*}

So Jem pays £69.50 in month 100.

(b)

解法一

思路

展开

用等差数列求和公式， $n=300$ 。

答题过程

展开

\begin{align*} S_{300} =&\,\frac{300}{2}\{2(20)+(300-1)(0.5)\}\\ =&\,150(40+149.5)\\ =&\,28425. \end{align*}

So Jem pays a total of £28425.

(c)

解法一

思路

展开

scheme B 是等比数列。由第 1 月和第 300 月求公比，再用等比数列前 $300$ 项和。

答题过程

展开

Let the common ratio for scheme B be $r$ .

Since the first payment is £20 and the 300th payment is £250,

\begin{align*} 20r^{299}=&\,250\\ r=&\,\big(\frac{250}{20}\big)^{1/299}. \end{align*}

\begin{align*} r=&\,1.008483031705\ldots . \end{align*}

Use the exact value $r=12.5^{1/299}$ in the following calculations, rather than its rounded decimal.

The total paid into scheme B is

\begin{align*} S_{300} =&\,\frac{20(1-r^{300})}{1-r}\\ =&\,27362.948\ldots . \end{align*}

The difference is

\begin{align*} 28425-27362.948\ldots =&\,1062.051\ldots . \end{align*}

To the nearest £10, the difference is

\begin{align*} \text{£}1060. \end{align*}

(b) 也可先求最后一项 $20+299(0.5)=169.50$ ，再用首末项公式：

\frac{300}{2}(20+169.50)=28425.

\log r=\frac{\log12.5}{299}.

这与前面的 $299$ 次根给出相同的未舍入公比。

P2 2024 Oct Q7

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

(c)

解法一

思路

答题过程