等差数列的首项为 $a$，公差为 $0.5$，其中 $a$ 是一个整数。求 $a$ 的最小值，使得前 36 项的和至少为 500。 数列 $u_0, u_1, u_2, \dots$ 的定义为： $u_0 = 400$ 且当 $n \ge 1$ 时 $u_n = 1.01u_{n-1} - x$，其中 $x$ 是一个整数。 证明 $u_n = 1.01^n(400) - 100x(1.01^n - 1)$。 已知 $x = 16$ 且 $y = 1.01u_k$（其中 $0 < y < 16$），求 $k$ 和 $y$ 的值。

1. Set up the inequality for the sum of the first 36 terms.

Using the formula for the sum of the first $n$ terms of an arithmetic sequence, $S_n = \frac{n}{2}(2a + (n-1)d)$, we substitute $n = 36$ and $d = 0.5$:
$S_{36} = \frac{36}{2}(2a + (36 - 1)(0.5))$
$S_{36} = 18(2a + 17.5)$
Given that the sum is at least 500, we set up the inequality:
$18(2a + 17.5) \ge 500$

2. Solve the linear inequality for $a$.

$36a + 315 \ge 500$
$36a \ge 185$
$a \ge \frac{185}{36}$
$a \ge 5\frac{5}{36}$

3. Determine the smallest integer value of $a$.

Since $a$ must be an integer, the smallest integer greater than or equal to $5\frac{5}{36}$ is 6.
$a = 6$

1. Use repeated substitution to find the first few terms and identify a pattern.

$u_1 = 1.01(400) - x$
$u_2 = 1.01u_1 - x$
$\phantom{u_2} = 1.01(1.01(400) - x) - x$
$\phantom{u_2} = 1.01^2(400) - x(1.01 + 1)$
$u_3 = 1.01u_2 - x$
$\phantom{u_3} = 1.01(1.01^2(400) - x(1.01 + 1)) - x$
$\phantom{u_3} = 1.01^3(400) - x(1.01^2 + 1.01 + 1)$

2. Generalise the expression for the $n$-th term.

By extending the observed pattern to $n$ steps, we obtain:
$u_n = 1.01^n(400) - x(1.01^{n-1} + 1.01^{n-2} + \dots + 1.01 + 1)$

3. Apply the sum formula for a geometric progression to simplify the expression.

The series in the bracket, $1 + 1.01 + \dots + 1.01^{n-1}$, is a geometric series with first term $a = 1$, common ratio $r = 1.01$, and $n$ terms.

$\text{Sum} = \frac{a(r^n - 1)}{r - 1}$
$\text{Sum} = \frac{1(1.01^n - 1)}{1.01 - 1}$
$\text{Sum} = \frac{1.01^n - 1}{0.01} = 100(1.01^n - 1)$

Substituting this sum back into the expression for $u_n$:
$u_n = 1.01^n(400) - x[100(1.01^n - 1)]$
$u_n = 1.01^n(400) - 100x(1.01^n - 1) \quad \text{(Shown)}$

1. Substitute $x=16$ into the given formula for $u_n$ to express $u_k$.

Given from (b)(i): $u_n = 1.01^n(400) - 100x(1.01^n - 1)$.
Substitute $n=k$ and $x=16$:
$u_k = 1.01^k(400) - 1600(1.01^k - 1)$
$u_k = 400(1.01^k) - 1600(1.01^k) + 1600$
$u_k = 1600 - 1200(1.01^k)$

2. Express $y$ in terms of $k$ and apply the given inequality $0 < y < 16$.

Since $y = 1.01 u_k$:
$y = 1.01(1600 - 1200(1.01^k))$
$y = 1616 - 1212(1.01^k)$

Applying $0 < y < 16$:
$0 < 1616 - 1212(1.01^k) < 16$

3. Solve the exponential inequality for $k$.

Subtract 1616:
$-1616 < -1212(1.01^k) < -1600$

Divide by $-1212$ (flip the inequality signs):
$\frac{1600}{1212} < 1.01^k < \frac{1616}{1212}$
$\frac{400}{303} < 1.01^k < \frac{4}{3}$

Taking natural logarithms on all parts:
$\frac{\ln(400/303)}{\ln 1.01} < k < \frac{\ln(4/3)}{\ln 1.01}$
$27.912 < k < 28.912$

4. State the integer value of $k$ and calculate the corresponding $y$.

Since $k$ represents an index in the sequence, it must be an integer.
Therefore, $k = 28$.

Substitute $k = 28$ back into the equation for $y$:
$y = 1616 - 1212(1.01^{28})$
$y \approx 14.595 \approx 14.6$ (to 3 s.f.)