2021-I-17 – Solving Master

Ans: (a)

6

(b)

33

Let $a$ and $d$ be the first term and the common difference of the arithmetic sequence respectively.
${\begin{cases} a + 4 d = 26 & \dots ① \\ a + 11 d = 61 & \dots ② \end{cases}$

$② - ①$ , we have

$\begin{array}{rcl} 7 d & = & 35 \\ d & = & 5 \end{array}$

Sub $d = 5$ into $①$ , we have

$\begin{array}{rcl} a + 4 (5) & = & 26 \\ a & = & 6 \end{array}$

$∴ A (1) = 6$ .
$\begin{array}{rcl} \log_{8} (G (1) G (2) G (3) \dots G (k)) & < & 999 \\ \frac{\log_{2} (G (1) G (2) G (3) \dots G (k))}{\log_{2} 8} & < & 999 \\ \frac{\log_{2} G (1) + \log_{2} G (2) + \dots + \log_{2} G (k)}{3} & < & 999 \\ A (1) + A (2) + \dots + A (k) & < & 2997 \\ \frac{k}{2} [2 (6) + (k - 1) 5] & < & 2997 \\ 5 k^{2} + 7 k - 5994 & < & 0 \end{array}$

$∴ - 35.330 766 67 < k < 33.930 766 67$ .

Therefore, the greatest value of $k$ is $33$ .