Ans: (a) $6$ (b) $33$
- Let $a$ and $d$ be the first term and the common difference of the arithmetic sequence respectively.
$\left\{ \begin{array}{ll}
a+4d = 26 & \ldots \unicode{x2460} \\
a+11d =61 & \ldots \unicode{x2461}
\end{array}\right.$$\unicode{x2461} -\unicode{x2460}$, we have
$\begin{array}{rcl}
7d & = & 35 \\
d & = & 5
\end{array}$Sub $d=5$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
a + 4(5) & = & 26 \\
a & = & 6
\end{array}$$\therefore A(1)=6$.
-
$\begin{array}{rcl}
\log_8\left( G(1) G(2) G(3) \ldots G(k) \right) & < & 999 \\ \dfrac{\log_2 \left( G(1) G(2) G(3) \ldots G(k) \right)}{\log_2 8} & < & 999 \\ \dfrac{\log_2 G(1) +\log_2 G(2) +\ldots +\log_2 G(k)}{3} & < & 999 \\ A(1) +A(2) +\ldots +A(k) & < & 2997 \\ \dfrac{k}{2}\left[ 2(6) +(k-1)5\right] & < & 2997 \\ 5k^2 +7k -5994 & < & 0 \\ \end{array}$$\therefore -35.330\ 766\ 67< k <33.930\ 766\ 67$.
Therefore, the greatest value of $k$ is $33$.
