2018-I-16

$\left\{ \begin{array}{ll}
ar^2 = 720 & \ldots \unicode{x2460} \\
ar^3=864 & \ldots \unicode{x2461}
\end{array}\right.$

$\unicode{x2461} \div \unicode{x2460}$, we have

$\begin{array}{rcl}
r & = & \dfrac{6}{5}\ \ldots \unicode{x2462}
\end{array}$

Sub. $\unicode{x2462}$ into $\unicode{x2460}$, we have

$\begin{array}{rcl}
a\left(\dfrac{6}{5}\right)^2 & = & 720 \\
a & = & 500
\end{array}$

Therefore, the 1st term is $500$.

$\begin{array}{rcl}
T(n+1) + T(2n+1) & < & 5 \times 10^{14} \\ 500\left(\dfrac{6}{5} \right)^n + 500\left(\dfrac{6}{5} \right)^{2n} & < & 5 \times 10^{14} \\ \left(\dfrac{6}{5}\right)^{2n} + \left(\dfrac{6}{5}\right)^n - 10^{12} & < & 0 \\ \left[\left(\dfrac{6}{5}\right)^n\right]^2 + \left(\dfrac{6}{5}\right)^n - 10^{12} & < & 0 \\ \end{array}$

Therefore, we have

$\dfrac{-1-\sqrt{1^2-4\times 1 \times (-10^{12})}}{2\times 1} < \left(\dfrac{6}{5}\right)^n < \dfrac{-1+\sqrt{1^2-4\times 1 \times (-10^{12})}}{2\times 1}$.

Since $\dfrac{6}{5} > 0$, then we have

$0 < \left(\dfrac{6}{5}\right)^n < \dfrac{-1+\sqrt{4 \times 10^{12} + 1}}{2\times 1}$.

Consider

$\begin{array}{rcl}
\left(\dfrac{6}{5}\right)^n & < & \dfrac{-1+\sqrt{4 \times 10^{12} + 1}}{2\times 1} \\ \log \left( \dfrac{6}{5} \right)^n & < & \log \dfrac{-1+\sqrt{4 \times 10^{12} + 1}}{2\times 1} \\ n \log \left( \dfrac{6}{5} \right) & < & \log \dfrac{-1+\sqrt{4 \times 10^{12} + 1}}{2\times 1} \\ n & < & 75.775\ 516\ 08 \end{array}$

Therefore, the greatest value of $n$ is $75$.