题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Given that, in a particular geometric series,

find, for this series,

(a) the common ratio,

(4)

(b) the first term.

(2)

解答

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设首项为 $a$ ，公比为 $r$ 。用 $S_3=70.2$ 和 $S_\infty=75$ 建两个方程，再消去 $a$ 。

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For a geometric series,

\begin{align*} S_3=&\,\frac{a(1-r^3)}{1-r},\\ S_\infty=&\,\frac{a}{1-r}. \end{align*}

\begin{align*} \frac{a(1-r^3)}{1-r}=&\,70.2,\\ \frac{a}{1-r}=&\,75. \end{align*}

Substitute $\frac{a}{1-r}=75$ into the first equation:

\begin{align*} 75(1-r^3)=&\,70.2\\ 1-r^3=&\,0.936\\ r^3=&\,0.064\\ r=&\,0.4. \end{align*}

Therefore, the common ratio is $0.4$ .

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把 $r=0.4$ 代回无穷和公式即可。

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Using

\begin{align*} \frac{a}{1-r}=75 \end{align*}

with $r=0.4$ ,

\begin{align*} \frac{a}{1-0.4}=&\,75\\ \frac{a}{0.6}=&\,75\\ a=&\,45. \end{align*}

So the first term is $45$ .