2020-I-16 – Solving Master

Ans: (a)

64

(b)

96

Let $a$ and $r$ be the first term and the common ratio of the geometric sequence respectively.
${\begin{cases} a r^{2} = 144 & \dots ① \\ a r^{5} = 486 & \dots ② \end{cases}$

$② \div ①$ , we have

$\begin{array}{rcl} r^{3} & = & \frac{27}{8} \\ r & = & \frac{3}{2} \end{array}$

Sub. $r = \frac{3}{2}$ into $①$ , we have

$\begin{array}{rcl} a (\frac{3}{2})^{2} & = & 144 \\ a & = & 64 \end{array}$

Therefore, the first term of the sequence is $64$ .
$\begin{array}{rcl} S (n) & > & 8 \times 10^{18} \\ \frac{64 [1 - (\frac{3}{2})^{n}]}{1 - \frac{3}{2}} & > & 8 \times 10^{18} \\ 128 [(\frac{3}{2})^{n} - 1] & > & 8 \times 10^{18} \\ (\frac{3}{2})^{n} & > & 6.25 \times 10^{16} + 1 \\ \log (\frac{3}{2})^{n} & > & \log (6.25 \times 10^{16} + 1) \\ n \log (\frac{3}{2}) & > & \log (6.25 \times 10^{16} + 1) \\ n & > & \frac{\log (6.25 \times 10^{16} + 1)}{\log (\frac{3}{2})} \\ n & > & 95.381 679 41 \end{array}$

Therefore, the least value of $n$ is $96$ .