一个等差数列的前 100 项之和为 10,000。该数列的第一项、第二项和第五项构成一个等比数列的三个连续项。求该等差数列的首项和非零公差。

1. Form an equation using the sum of the first 100 terms of the arithmetic progression.

Let the first term be $a$ and the common difference be $d$.
Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:
$S_{100} = \frac{100}{2}[2a + 99d] = 10000$
$50(2a + 99d) = 10000$
$2a + 99d = 200 \quad \text{--- (1)}$

2. Identify the $1^\text{st}$, $2^\text{nd}$ and $5^\text{th}$ terms of the AP.

The respective terms in the AP are:
$T_1 = a$
$T_2 = a + d$
$T_5 = a + 4d$

3. Form an equation using the property of the consecutive terms of a geometric progression.

Since $T_1$, $T_2$ and $T_5$ are three consecutive terms of a GP, they share a common ratio:
$\frac{T_2}{T_1} = \frac{T_5}{T_2}$
$\frac{a+d}{a} = \frac{a+4d}{a+d}$

4. Simplify the GP equation to find the relationship between $a$ and $d$.

Cross-multiplying gives:
$(a+d)^2 = a(a+4d)$
$a^2 + 2ad + d^2 = a^2 + 4ad$
$d^2 - 2ad = 0$
$d(d - 2a) = 0$
Since the question states the common difference is non-zero ($d \neq 0$):
$d = 2a \quad \text{--- (2)}$

5. Solve the simultaneous equations to find $a$ and $d$.

Substitute equation (2) into equation (1):
$2a + 99(2a) = 200$
$2a + 198a = 200$
$200a = 200 \implies a = 1$
Substitute $a = 1$ back into equation (2):
$d = 2(1) = 2$