The ninth, fifth and second terms of an arithmetic progression are successive terms of a geometric progression with first term $a$ and common ratio $r$, where $r \ne 1$ and $a > 0$. Find the value of $r$ and deduce that the geometric series is convergent. Using the value of $r$ found in (i), find the least value of $n$ such that the sum of all the terms after the $n^{th}$ term of the geometric progression is less than 1% of its sum of the first $n$ terms. The sum, $S_n$, of the first $n$ terms of another sequence is given by $S_n = n \ln\left(2^{q(n+1)}\right)$, where $q$ is a constant. Prove that the sequence follows an arithmetic progression.

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.