The second term of a geometric sequence is $48$ and the fourth term is $3$, find $r$

Question

I am struggling with this maths problem as I can't see the leap of logic that is used to get from knowing that $ar=48$ and $ar^3=3$ to knowing that one value of $r$ is $-\frac{1}{4}$ as given in the question.

I am pretty sure it's a simultaneous equations problem but I can't see what steps I should be taking. The mark scheme doesn't help much either: Mark scheme

I realise that this will be a trivial question for most but I would value a good explanation :)

Answer 1

$ar^3=3$ $ar=48$

Dividing these equations, $r^2=\frac 1{16}$

$r=\pm \frac 14$

Answer 2

If $a,48,b,3$ are in Geometric Series,

$$\dfrac{48}a=\dfrac b{48}=\dfrac3b$$

$$\implies b^2=48\cdot3\iff b=\pm12$$

$$\dfrac{48}a=\dfrac b{48}\implies a=\dfrac{48^2}b=\cdots$$

Answer 3

$ar = 48$

$\implies$ a = $\dfrac{48}{r}$

$ar^3 = 3$

$\implies$ a = $\dfrac{3}{r^3}$

So we say:

$a = a$

$\dfrac{48}{r}$ = $\dfrac{3}{r^3}$

$48r^3 = 3r$

$48r^2 = 3$

$r^2$ = $\dfrac{1}{16}$

=> r = $\pm \dfrac{1}{4}$