If three distinct numbers $$a$$, $$b$$, $$c$$ are in G.P. and the equations $$ax^{2} + 2bx + c = 0$$ and $$dx^{2} + 2ex + f = 0$$ have a common root, then which one of the following statements is correct?

We are told that the three non-zero and distinct numbers $$a$$, $$b$$, $$c$$ are in geometric progression (G.P.).

When three numbers are in G.P. we always write them as

$$b = ar \quad\text{and}\quad c = ar^{2},$$

where $$r \neq 1$$ is the common ratio. (Because the numbers are distinct, $$r$$ cannot be $$1$$.)

The first quadratic equation is

$$ax^{2}+2bx+c = 0.$$

Substituting $$b = ar$$ and $$c = ar^{2}$$ we obtain

$$a x^{2} + 2(ar)x + ar^{2} = 0.$$

Now we divide every term by $$a\;( \neq 0 )$$ to simplify:

$$x^{2} + 2rx + r^{2} = 0.$$

Next we examine the nature of this quadratic. Its discriminant is

$$\Delta = (2r)^{2} - 4(1)(r^{2}) = 4r^{2} - 4r^{2} = 0.$$

Because the discriminant is zero, the quadratic has a repeated (double) root. Using the standard relation “for $$x^{2}+2px+p^{2}=0$$ the root is $$x=-p$$”, we get

$$x = -r.$$

Thus the first equation has the single root

$$\alpha = -r.$$

The statement of the question says that the two quadratics

$$ax^{2}+2bx+c = 0 \quad\text{and}\quad dx^{2}+2ex+f = 0$$

have a common root. Therefore that common root must be $$\alpha = -r$$. Hence $$x=-r$$ must satisfy the second equation, giving

$$d(-r)^{2} + 2e(-r) + f = 0.$$

Simplifying term by term we have

$$dr^{2} - 2er + f = 0.$$

We now want to investigate the three numbers

$$\frac{d}{a},\quad\frac{e}{b},\quad\frac{f}{c}.$$

Remember that $$b = ar$$ and $$c = ar^{2}$$, so

$$\frac{e}{b} = \frac{e}{ar},\quad\frac{f}{c} = \frac{f}{ar^{2}}.$$

To show that these three numbers are in arithmetic progression (A.P.) we must verify the defining relation for an A.P.:

$$2\left(\frac{e}{b}\right) = \frac{d}{a} + \frac{f}{c}.$$

Let us compute each side separately and then compare.

Left side:

$$2\!\left(\frac{e}{b}\right) = 2\!\left(\frac{e}{ar}\right) = \frac{2e}{ar}.$$

Right side:

$$\frac{d}{a} + \frac{f}{c} = \frac{d}{a} + \frac{f}{ar^{2}}.$$

To compare them conveniently, we eliminate denominators by multiplying every term by $$ar^{2}$$:

Left side × $$ar^{2}$$:

$$\frac{2e}{ar}\; \cdot ar^{2} = 2er.$$

Right side × $$ar^{2}$$:

$$\left(\frac{d}{a}\right)ar^{2} + \left(\frac{f}{ar^{2}}\right)ar^{2} = dr^{2} + f.$$

Hence the A.P. condition becomes the equality

$$2er = dr^{2} + f.$$

But this is exactly the relation we obtained earlier from the common-root condition $$dr^{2} - 2er + f = 0$$, because that equation rearranges to

$$dr^{2} + f = 2er.$$

Since the required equality holds, the three numbers $$\dfrac{d}{a},\ \dfrac{e}{b},\ \dfrac{f}{c}$$ are indeed in arithmetic progression. None of the other options can be justified by the given information.

Hence, the correct answer is Option A.