Three numbers are in an increasing geometric progression with common ratio $$r$$. If the middle number is doubled, then the new numbers are in an arithmetic progression with common difference $$d$$. If the fourth term of GP is $$3r^2$$, then $$r^2 - d$$ is equal to:

Let the three numbers in the increasing GP be $$a, ar, \text{ and } ar^2$$.

• Since it is an increasing progression, we know $$r > 1$$ and $$a > 0$$.

According to the problem, if the middle number is doubled, the new sequence $$a, 2ar, ar^2$$ forms an AP.

In an AP, twice the middle term equals the sum of the outer terms:

$$2(2ar) = a + ar^2$$

Dividing by $$a$$ (since $$a \neq 0$$):

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

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

Solving for $$r$$ using the quadratic formula:

$$r = \frac{4 \pm \sqrt{16 - 4}}{2} = \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$$

Since the GP is increasing ($$r > 1$$), we take $$r = 2 + \sqrt{3}$$.

We are given that the fourth term ($$ar^3$$) is equal to $$3r^2$$:

$$ar^3 = 3r^2$$

$$ar = 3$$

The common difference $$d$$ of the AP ($$a, 2ar, ar^2$$) is:

$$d = 2ar - a$$

Substitute $$ar = 3$$:

$$d = 2(3) - a = 6 - a$$

To find $$a$$, use $$ar = 3$$:

$$a = \frac{3}{r} = \frac{3}{2 + \sqrt{3}} = 3(2 - \sqrt{3}) = 6 - 3\sqrt{3}$$

Now calculate $$d$$:

$$d = 6 - (6 - 3\sqrt{3}) = 3\sqrt{3}$$

First, find $$r^2$$:

$$r^2 = (2 + \sqrt{3})^2 = 4 + 3 + 4\sqrt{3} = 7 + 4\sqrt{3}$$

Now, subtract $$d$$:

$$r^2 - d = (7 + 4\sqrt{3}) - 3\sqrt{3}$$

$$r^2 - d = 7 + \sqrt{3}$$

The value of $$r^2 - d$$ is $$7 + \sqrt{3}$$, which corresponds to option (D).