Three numbers are in an increasing geometric progression with...

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

Let first term is $a$

$⇒ \dfrac{a}{r}, a, ar \in G . P$

Given that if middle term is doubled

$⇒ \dfrac{a}{r}, 2a, ar \in A . P$

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

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

${r}^{2} + 1 = 4r$

${r}^{2} - 4r + 1 = 0$

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

$⇒ r = \dfrac{4 \pm \sqrt{12}}{2} = 2 + \sqrt{3}, 2 - \sqrt{3}$

But it is an increasing $G . P$

$∴ r = 2 + \sqrt{3}$

Given that Fourth term of $G . P ⇒ {t}_{4} = a{r}^{2} = 3{r}^{2}$

$⇒ a = 3 .$

Common difference of an $A . P = d = {t}_{2} - {t}_{1}$

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

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

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

$= 7 + \sqrt{3}$