-
$\left\{ \begin{array}{ll}
\beta = 5 \alpha – 18 & \ldots \unicode{x2460} \\
\beta = \alpha^2 -13 \alpha + 63 & \ldots \unicode{x2461}
\end{array} \right.$把 $\unicode{x2460}$ 代入 $\unicode{x2461}$,可得
$\begin{array}{rcl}
5 \alpha – 18 & = & \alpha^2 – 13 \alpha + 63 \\
\alpha^2 – 18 \alpha + 81 & = & 0 \\
(\alpha – 9)^2 & = & 0
\end{array}$所以,$\alpha = 9$ (重根)。
把 $\alpha = 9$ 代入 $\unicode{x2460}$,可得
$\begin{array}{rcl}
\beta & = & 5(9) – 18 \\
\beta & = & 27
\end{array}$所以,$\beta = 27$。
- 留意該公差
$\begin{array}{cl}
= & \log \beta – \log \alpha \\
= & \log \dfrac{\beta}{\alpha} \\
= & \log \dfrac{27}{9} \\
= & \log 3
\end{array}$設 $S(n)$ 為該數列首 $n$ 項之和。
$\begin{array}{rcl}
S(n) & > & 888 \\
\dfrac{n}{2}[2\log 9 + (n – 1) \log 3] & > & 888 \\
2n\log 9 + n^2 \log 3 – n \log 3 & > & 1776 \\
n^2 \log 3 +2n\log 3^2 – n \log 3 – 1776 & > & 0 \\
n^2 \log 3 + 4n\log 3 – n \log 3 – 1776 & > & 0 \\
n^2 \log 3 + 3n \log 3 – 1776 & < & 0 \end{array}$所以,$n < - 62.529\ 289\ 81$ 或 $n > 59.529\ 289\ 81$。
由此,$n$ 的最小值為 $60$。
2019-I-16
答案:(a) $\alpha = 9$, $\beta = 27$ (b) $60$