一个等差数列的第九项、第五项和第二项是一个首项为 $a$、公比为 $r$ 的等比数列的连续三项，其中 $r \ne 1$ 且 $a > 0$。求 $r$ 的值并推导证明该等比级数是收敛的。 利用在 (i) 中求得的 $r$ 值，求 $n$ 的最小值，使得该等比数列在第 $n$ 项之后的所有项之和小于其前 $n$ 项之和的 1%。 另一个数列的前 $n$ 项和 $S_n$ 由 $\displaystyle S_n = n\ln 2^{q(n+1)}$ 给出，其中 $q$ 是一个常数。证明该数列是一个等差数列。

1. Set up equations relating the terms of the arithmetic and geometric progressions.

Let the arithmetic progression have a first term $A$ and a common difference $D$.
The $9^{\text{th}}$, $5^{\text{th}}$ and $2^{\text{nd}}$ terms of the AP correspond to successive terms of the GP (which we can denote as starting with $a$, followed by $ar$ and $ar^2$):
$A + 8D = a$
$A + 4D = ar$
$A + D = ar^2$

2. Eliminate $A$ and $D$ to solve for $r$.

Subtracting the second equation from the first to eliminate $A$:
$(A + 8D) - (A + 4D) = a - ar$
$4D = a(1 - r) \quad \text{--- (1)}$

Subtracting the third equation from the second:
$(A + 4D) - (A + D) = ar - ar^2$
$3D = ar(1 - r) \quad \text{--- (2)}$

Dividing equation (1) by equation (2):
$\frac{4D}{3D} = \frac{a(1 - r)}{ar(1 - r)}$

Since $a > 0$ and $r \ne 1$, $a(1-r) \ne 0$. We can cancel out the common terms:
$\frac{4}{3} = \frac{1}{r}$
$r = \frac{3}{4}$

3. Deduce the convergence of the geometric series.

For an infinite geometric series to be convergent, the common ratio $r$ must satisfy $|r| < 1$.
Since $r = \frac{3}{4}$, we have $\left| \frac{3}{4} \right| < 1$.
Thus, the geometric series is convergent.

1. Derive the common ratio $r$ from part (a)(i) as required for this part.

From part (a)(i), let the arithmetic progression have first term $A$ and common difference $D$.
$T_9 = A + 8D = a$
$T_5 = A + 4D = ar$
$T_2 = A + D = ar^2$
Subtracting the equations yields:
$4D = a - ar = a(1-r)$
$3D = ar - ar^2 = ar(1-r)$
Dividing the two expressions gives:
$\frac{4}{3} = \frac{1}{r} \implies r = \frac{3}{4}$

2. Formulate the inequality comparing the sum after the $n^{\text{th}}$ term to the sum of the first $n$ terms.

The sum of all terms after the $n^{\text{th}}$ term is an infinite geometric series starting with $ar^n$:
$S_\infty - S_n = \frac{a}{1-r} - \frac{a(1-r^n)}{1-r} = \frac{ar^n}{1-r}$
The sum of the first $n$ terms is $S_n = \frac{a(1-r^n)}{1-r}$.
Given that the sum after the $n^{\text{th}}$ term is less than $1\%$ of $S_n$:
$\frac{ar^n}{1-r} < 0.01 \left( \frac{a(1-r^n)}{1-r} \right)$

3. Simplify the inequality and substitute the value of $r$.

Since $a > 0$ and $r = \frac{3}{4} < 1$, the term $\frac{a}{1-r}$ is strictly positive. We can divide both sides by it without changing the inequality sign:
$r^n < 0.01(1 - r^n)$
$r^n + 0.01r^n < 0.01$
$1.01r^n < 0.01 \implies r^n < \frac{1}{101}$
Substitute $r = 0.75$:
$0.75^n < \frac{1}{101}$

4. Solve for $n$ using logarithms to find the least integer value.

Take the natural logarithm on both sides:
$n \ln 0.75 < \ln \left(\frac{1}{101}\right)$
Since $\ln 0.75$ is negative, dividing by it reverses the inequality sign:
$n > \frac{\ln(1/101)}{\ln 0.75}$
$n > 16.04$
Since $n$ must be an integer, the least value of $n$ is $17$.

1. Simplify the expression for the sum to $n$ terms, $S_n$.

$S_n = n\ln 2^{q(n+1)} = n \cdot q(n+1)\ln 2 = q(n^2 + n)\ln 2$

2. Use the relationship $u_n = S_n - S_{n-1}$ to find the general $n^{\text{th}}$ term of the sequence for $n \ge 2$.

$\begin{aligned} u_n &= S_n - S_{n-1} \\ &= q(n^2 + n)\ln 2 - q((n-1)^2 + (n-1))\ln 2 \end{aligned}$

3. Simplify the expression for $u_n$.

$\begin{aligned} u_n &= q\ln 2 [n^2 + n - (n^2 - 2n + 1 + n - 1)] \\ &= q\ln 2 [n^2 + n - (n^2 - n)] \\ &= 2qn\ln 2 \end{aligned}$

4. Show that the difference between consecutive terms is a constant to prove it is an arithmetic progression.

$\begin{aligned} u_n - u_{n-1} &= 2qn\ln 2 - 2q(n-1)\ln 2 \\ &= 2q\ln 2(n - n + 1) \\ &= 2q\ln 2 \end{aligned}$
(Note: For $n=1$, $u_1 = S_1 = 1\ln 2^{2q} = 2q\ln 2$, which perfectly follows the formula $u_n = 2qn\ln 2$.)

Since $u_n - u_{n-1} = 2q\ln 2$ is a constant independent of $n$ for all $n \ge 2$, the sequence follows an arithmetic progression.