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 $r$ 的等比数列的连续项，其中 $r \ne 1$ 且 $a > 0$。求 $r$ 的值并推导说明该等比级数是收敛的。 利用在 (i) 中求得的 $r$ 的值，求最小的 $n$ 值，使得该等比数列在第 $n$ 项之后的所有项之和小于其前 $n$ 项和的 1%。 另一个数列的前 $n$ 项和 $S_n$ 由 $S_n = n \ln\left(2^{q(n+1)}\right)$ 给出，其中 $q$ 为常数。证明该数列是等差数列。

1. Express the 9th, 5th, and 2nd terms of the arithmetic progression in terms of its first term and common difference, and equate them to the terms of the geometric progression.

Let the arithmetic progression have first term $A$ and common difference $d$. The successive terms of the geometric progression are $a$, $ar$, and $ar^2$.
$A + 8d = a$
$A + 4d = ar$
$A + d = ar^2$

2. Eliminate the first term $A$ by finding the differences between the equations.

Subtracting the equations to eliminate $A$:
$(A + 8d) - (A + 4d) = a - ar \implies 4d = a(1 - r)$
$(A + 4d) - (A + d) = ar - ar^2 \implies 3d = ar(1 - r)$

3. Solve for the common ratio $r$ by dividing the two expressions for $d$.

Dividing the second difference by the first difference:
$\frac{3d}{4d} = \frac{ar(1 - r)}{a(1 - r)}$
Since $r \neq 1$ and $a > 0$, we have $a(1 - r) \neq 0$. Also, if $d = 0$, then $a = ar \implies r = 1$ (which is rejected), so $d \neq 0$.
Dividing out the common terms gives:
$\frac{3}{4} = r$

4. Deduce that the geometric series is convergent using the value of $r$.

For a geometric series to be convergent, the common ratio $r$ must satisfy $|r| < 1$.
Since $r = \frac{3}{4}$, we have $\left|\frac{3}{4}\right| < 1$, which satisfies the condition. Therefore, the geometric series is convergent.

1. Find the value of the common ratio $r$ as required from part (i).

Let the arithmetic progression have first term $A$ and common difference $D$.
We are given $T_9 = A + 8D = a$, $T_5 = A + 4D = ar$, and $T_2 = A + D = ar^2$.
Then $T_9 - T_5 = a(1-r) = 4D$ and $T_5 - T_2 = ar(1-r) = 3D$.
Dividing the two expressions yields:
$\frac{a(1-r)}{ar(1-r)} = \frac{4D}{3D} \implies \frac{1}{r} = \frac{4}{3} \implies r = \frac{3}{4}$

2. Formulate the inequality given in the problem for part (ii).

The sum of all terms after the $n^{th}$ term is an infinite geometric series starting with the $(n+1)^{th}$ term $ar^n$. Its sum is $\frac{ar^n}{1-r}$.
Alternatively, this is $S_\infty - S_n = \frac{a}{1-r} - \frac{a(1-r^n)}{1-r} = \frac{ar^n}{1-r}$.
The condition given is that this sum 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 to isolate $r^n$.

Since $a > 0$ and $1-r = 1 - \frac{3}{4} > 0$, we can divide both sides by $\frac{a}{1-r}$ without changing the inequality sign:
$r^n < 0.01(1 - r^n)$
$r^n < 0.01 - 0.01r^n$
$1.01r^n < 0.01$
$r^n < \frac{0.01}{1.01} = \frac{1}{101}$

4. Substitute the value of $r$ and solve for $n$ using logarithms.

Substitute $r = 0.75$:
$0.75^n < \frac{1}{101}$
Taking the natural logarithm on both sides:
$n \ln(0.75) < \ln\left(\frac{1}{101}\right)$
Since $\ln(0.75)$ is negative, the inequality sign flips when dividing:
$n > \frac{\ln(1/101)}{\ln(0.75)}$
$n > 16.043$
Since $n$ must be an integer, the least value of $n$ is $17$.

1. Simplify the expression for $S_n$ using the power rule for logarithms.

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

2. Set up an expression for the $n$-th term, $u_n$, using the relationship $u_n = S_n - S_{n-1}$.

$u_n = S_n - S_{n-1}$
$u_n = (q \ln 2)(n^2 + n) - (q \ln 2)\left((n-1)^2 + (n-1)\right)$

3. Simplify the expression for $u_n$.

$u_n = (q \ln 2) \left[ n^2 + n - (n^2 - 2n + 1 + n - 1) \right]$
$u_n = (q \ln 2) (n^2 + n - n^2 + n)$
$u_n = 2n q \ln 2$

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

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