Find the first negative coefficient in the expansion of $(4 + 3x)^{5/2}$ in a series of ascending powers of $x$, where $|x| < \dfrac{4}{3}$. Give your answer as a fraction in its lowest terms.

1. Factor out $4$ to apply the standard expansion $(1+y)^n$.

$(4 + 3x)^{5/2} = \left[4 \left(1 + \frac{3}{4}x\right)\right]^{5/2} = 4^{5/2} \left(1 + \frac{3}{4}x\right)^{5/2} = 32 \left(1 + \frac{3}{4}x\right)^{5/2}$

2. Identify the term which will have the first negative coefficient.

The coefficient of $x^r$ involves the product $n(n-1)(n-2)\dots(n-r+1)$.
For $n = \frac{5}{2}$, the factors are:
$r=1: \frac{5}{2}$
$r=2: \frac{3}{2}$
$r=3: \frac{1}{2}$
$r=4: -\frac{1}{2}$
The first negative factor appears when $r=4$. Therefore, the first negative coefficient corresponds to the term in $x^4$.

3. Calculate the coefficient of $x^4$.

The coefficient of $x^4$ is:
$32 \times \frac{\left(\frac{5}{2}\right)\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{4!} \times \left(\frac{3}{4}\right)^4$
$= 32 \times \left(-\frac{15}{16 \times 24}\right) \times \frac{81}{256}$
$= 32 \times \left(-\frac{5}{128}\right) \times \frac{81}{256}$
$= -\frac{5}{4} \times \frac{81}{256} = -\frac{405}{1024}$