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Step-by-Step Solution
Step 1: Recognize the Given Conditions
We are given three distinct numbers $a$, $b$, $c$ in geometric progression (G.P.), which means:
$$
b^2 = ac.
$$
We also know there are two quadratic equations:
$$
ax^2 + 2bx + c = 0 \quad \text{and} \quad dx^2 + 2ex + f = 0
$$
that share a common root.
Step 2: Analyze the First Quadratic Equation
Consider the first equation:
$$
ax^2 + 2bx + c = 0.
$$
The discriminant $\Delta$ of this equation is given by
$$
\Delta = (2b)^2 - 4 \cdot a \cdot c
= 4b^2 - 4ac.
$$
Since $b^2 = ac$, it follows that:
$$
4b^2 - 4ac = 4ac - 4ac = 0.
$$
Thus, $\Delta = 0$, which tells us the roots of this equation are real and equal. Let both equal roots be $\alpha$.
Step 3: Find the Common Root
If $\alpha$ is the (repeated) root of the first equation, then by Viète's formula for the sum of roots,
$$
\alpha + \alpha = -\frac{2b}{a} \;\;\Longrightarrow\;\; 2\alpha = -\frac{2b}{a} \;\;\Longrightarrow\;\; \alpha = -\frac{b}{a}.
$$
Because the two quadratics share a common root, $x = -\frac{b}{a}$ must also satisfy
$$
dx^2 + 2ex + f = 0.
$$
Step 4: Substitute the Common Root in the Second Equation
Substitute $x = -\tfrac{b}{a}$ into the second equation:
$$
d\left(-\frac{b}{a}\right)^2 + 2e\left(-\frac{b}{a}\right) + f = 0.
$$
Simplify:
$$
d\,\frac{b^2}{a^2} \;-\; 2\,\frac{eb}{a} \;+\; f \;=\; 0.
$$
Multiply through by $a^2$ to clear denominators:
$$
d\,b^2 - 2\,a\,e\,b + f\,a^2 = 0.
$$
But since $b^2 = ac$, replace $b^2$ with $ac$:
$$
d\,a\,c - 2\,a\,e\,b + f\,a^2 = 0
\quad\Longrightarrow\quad
dc - 2eb + fa = 0.
$$
Step 5: Express the Result to Reveal an Arithmetic Progression
Divide the entire equation $dc - 2eb + fa = 0$ by $ac$ (using $b^2 = ac$, we may also switch to that wherever needed) carefully to rewrite each term as a fraction of $a,b,c$:
$$
\frac{dc}{ac} \;-\; \frac{2eb}{ac} \;+\; \frac{fa}{ac} = 0
\;\;\Longrightarrow\;\;
\frac{d}{a} - \frac{2e}{b} + \frac{f}{c} = 0.
$$
Reorganize:
$$
\frac{2e}{b} = \frac{d}{a} + \frac{f}{c}.
$$
This implies that
$$
\frac{d}{a}, \quad \frac{e}{b}, \quad \frac{f}{c}
$$
form three terms of an arithmetic progression (A.P.), because the middle term $\frac{e}{b}$ is the average of the other two.
Conclusion
Hence, the correct statement is:
$$
\frac{d}{a}, \quad \frac{e}{b}, \quad \frac{f}{c}
\text{ are in A.P.}
$$
