Three numbers are in an increasing geometric progression with common ratio $r.$

Let first term is $a$

$\Rightarrow \frac{a}{r},a,ar \in G.P$ 

Given that if middle term is doubled 

 $\Rightarrow \frac{\mathrm{a}}{\mathrm{r}},2\mathrm{a},ar \in A.P$ 

If $a, b, c \in A. P \Rightarrow 2b =a + c$
$4a=ar+\frac{a}{r}$

$4=r+\frac{1}{r}$

$r^2+1=4r$

$r^2-4r+1=0$ 

If$a{x}^2 + bx +c = 0 \Rightarrow x = \frac{-b \pm \sqrt{b^2-4ac}}{2ac}$

$\Rightarrow r=\frac{4\pm \sqrt{12}}{2}=2+\sqrt{3},2-\sqrt{3}$

But it is an increasing $G. P$

$\therefore r = 2 + \sqrt{3}$

Given that Fourth term of  $G.P \Rightarrow t_4 = a{r}^2 = 3r^2$

$\Rightarrow a = 3.$

Common difference of an $A.P = d = t_2 - t_1$

$d=2a-\frac{a}{r}=a\left(2-\frac{1}{r}\right)=3\left(2-\frac{1}{2+\sqrt{3}}\right)$

$=3(2-2+\sqrt{3})=3\sqrt{3}$
Hence $r^2-d=(2+\sqrt{3}{)}^2-(3\sqrt{3})$

$= 4 + 3 + 4\sqrt{3} - 3\sqrt{3}$

$=7+\sqrt{3}$