2019-I-16

Sub. $\unicode{x2460}$ into $\unicode{x2461}$, we have

$\begin{array}{rcl}
5 \alpha – 18 & = & \alpha^2 – 13 \alpha + 63 \\
\alpha^2 – 18 \alpha + 81 & = & 0 \\
(\alpha – 9)^2 & = & 0
\end{array}$

Therefore, $\alpha = 9$ (repeated).

Sub. $\alpha = 9$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
\beta & = & 5(9) – 18 \\
\beta & = & 27
\end{array}$

Therefore, $\beta = 27$.

$\begin{array}{cl}
= & \log \beta – \log \alpha \\
= & \log \dfrac{\beta}{\alpha} \\
= & \log \dfrac{27}{9} \\
= & \log 3
\end{array}$

Let $S(n)$ be the sum of the first $n$ terms of the sequence.

$\begin{array}{rcl}
S(n) & > & 888 \\
\dfrac{n}{2}[2\log 9 + (n – 1) \log 3] & > & 888 \\
2n\log 9 + n^2 \log 3 – n \log 3 & > & 1776 \\
n^2 \log 3 +2n\log 3^2 – n \log 3 – 1776 & > & 0 \\
n^2 \log 3 + 4n\log 3 – n \log 3 – 1776 & > & 0 \\
n^2 \log 3 + 3n \log 3 – 1776 & < & 0 \end{array}$

Therefore, $n < - 62.529\ 289\ 81$ or $n > 59.529\ 289\ 81$.

Hence, the least value of $n$ is $60$.