A sequence of numbers $u_{1}, u_{2}, u_{3}, \ldots$ has sum $S_{n}$ where $S_{n} = \sum_{r=1}^{n} u_{r}$. It is given that $S_{n} = 18 - \dfrac{2^{n+1}}{3^{n-2}}$.

1. Use the relationship $u_n = S_n - S_{n-1}$ to express the $n$th term.

$u_n = S_n - S_{n-1}$
$u_n = \left( 18 - \frac{2^{n+1}}{3^{n-2}} \right) - \left( 18 - \frac{2^{(n-1)+1}}{3^{(n-1)-2}} \right)$
$u_n = -\frac{2^{n+1}}{3^{n-2}} + \frac{2^n}{3^{n-3}}$

2. Factorise and simplify the expression for $u_n$.

$u_n = \frac{2^n}{3^{n-3}} - \frac{2^{n+1}}{3^{n-2}}$
Express with a common denominator of $3^{n-2}$:
$u_n = \frac{3(2^n)}{3^{n-2}} - \frac{2(2^n)}{3^{n-2}}$
$u_n = \frac{2^n(3 - 2)}{3^{n-2}}$
$u_n = \frac{2^n}{3^{n-2}}$

3. Show that the ratio of consecutive terms is a constant.

$\frac{u_{n+1}}{u_n} = \frac{ \frac{2^{n+1}}{3^{n-1}} }{ \frac{2^n}{3^{n-2}} }$
$\frac{u_{n+1}}{u_n} = \left( \frac{2^{n+1}}{2^n} \right) \left( \frac{3^{n-2}}{3^{n-1}} \right)$
$\frac{u_{n+1}}{u_n} = 2^1 \times 3^{-1} = \frac{2}{3}$
Since $\frac{u_{n+1}}{u_n}$ is a constant (independent of $n$), the sequence is a geometric progression.

1. Calculate the sum of the first 9 terms of the geometric sequence.

$\sum_{r=1}^{9} u_{r} = S_9 = 18 - \frac{2^{9+1}}{3^{9-2}}$
$S_9 = 18 - \frac{2^{10}}{3^7} = 18 - \frac{1024}{2187} = \frac{38342}{2187}$

2. Express the sum of the first 9 terms of the arithmetic progression in terms of the common difference, $d$.

$S_{AP} = \frac{n}{2} [2a + (n-1)d]$
$S_{AP} = \frac{9}{2} [2(-4) + (9-1)d]$
$S_{AP} = \frac{9}{2} [-8 + 8d] = -36 + 36d$

3. Equate the two sums and solve for $d$.

$-36 + 36d = \frac{38342}{2187}$
$36d = 36 + \frac{38342}{2187}$
$36d = \frac{78732 + 38342}{2187} = \frac{117074}{2187}$
$d = \frac{117074}{36 \times 2187} = \frac{58537}{39366} \approx 1.49 \text{ (to 3 s.f.)}$