等比數列的第 $n$ 項公式為 $u_n = a r^{n-1}$。

1. 建立方程：
   • 第 3 項為 $u_3 = a r^2 = 20$。
   • 第 5 項為 $u_5 = a r^4 = 12.8$。
2. 消去首項求公比： 將兩式相除以消去 $a$： $$\frac{u_5}{u_3} = \frac{a r^4}{a r^2} = r^2 = \frac{12.8}{20} = 0.64$$ 由於已知公比 $r > 0$，所以對 $0.64$ 開平方根即可證得 $r = 0.8$。

The $n$-th term of a geometric sequence is given by:

Given the 3rd term is $20$ and the 5th term is $12.8$, we have:

Dividing equation (2) by equation (1) to eliminate $a$:

Since $r > 0$, taking the positive square root:

利用 (a) 小題求出的 $r = 0.8$ 以及第 3 項的表達式 $a r^2 = 20$，代入求得首項 $a$ 的值。

From part (a), we have $r = 0.8$ and $ar^2 = 20$.

Substituting $r = 0.8$ into the equation:

1. 列出不等式：等比數列的前 $n$ 項和公式為 $S_n = \frac{a(1-r^n)}{1-r}$。 將 $a = 31.25, r = 0.8$ 代入公式，並令 $S_n > 156$： $$\frac{31.25(1-0.8^n)}{1-0.8} > 156$$
2. 代数化简： $$156.25(1-0.8^n) > 156 \implies 1-0.8^n > 0.9984 \implies 0.8^n < 0.0016$$
3. 對數求解：兩邊取自然對數（或常用對數）。注意由於底數 $0.8 < 1$，對數值為負數，兩邊同除以 $\log 0.8$ 時不等號方向需要翻轉： $$n > \frac{\log 0.0016}{\log 0.8}$$ 求出數值界限後，取大於該界限的最小整數即可。

The sum of the first $n$ terms $S_n$ of a geometric progression is:

Substitute $a = 31.25$ and $r = 0.8$ into the inequality $S_n > 156$:

Taking the common logarithm (base 10) of both sides:

Since $\log(0.8) \approx -0.0969 < 0$, dividing by $\log(0.8)$ reverses the inequality sign:

Since $n$ must be an integer, the smallest possible value of $n$ is: