Ans: (a) $64$ (b) $96$
- Let $a$ and $r$ be the first term and the common ratio of the geometric sequence respectively.
$\left\{\begin{array}{ll}
ar^2 = 144 & \ldots \unicode{x2460} \\
ar^5 = 486 & \ldots \unicode{x2461}
\end{array}\right.$$\unicode{x2461} \div \unicode{x2460}$, we have
$\begin{array}{rcl}
r^3 & = & \dfrac{27}{8} \\
r & = & \dfrac{3}{2}
\end{array}$Sub. $r = \dfrac{3}{2}$ into $\unicode{x2460}$, we have
$\begin{array}{rcl}
a(\dfrac{3}{2})^2 & = & 144 \\
a & = & 64
\end{array}$Therefore, the first term of the sequence is $64$.
-
$\begin{array}{rcl}
S(n) & > & 8 \times 10^{18} \\
\dfrac{64[1-(\frac{3}{2})^n]}{1-\frac{3}{2}} & > & 8 \times 10^{18} \\
128[ (\dfrac{3}{2})^n – 1] & > & 8 \times 10^{18} \\
(\dfrac{3}{2})^n & > & 6.25 \times 10^{16} + 1 \\
\log (\dfrac{3}{2} )^n & > & \log (6.25 \times 10^{16} + 1) \\
n\log (\dfrac{3}{2} ) & > & \log (6.25 \times 10^{16} + 1) \\
n & > & \dfrac{\log (6.25\times 10^{16} + 1)}{ \log (\frac{3}{2} ) } \\
n & > & 95.381\ 679\ 41
\end{array}$Therefore, the least value of $n$ is $96$.
