Key Concepts and Formulas:

1. Geometric Progression (G.P.): Three distinct numbers $A, B, C$ are in G.P. if $\frac{B}{A} = \frac{C}{B}$, which implies $B^2 = AC$.
2. Quadratic Equation Roots: For a quadratic equation $Px^2 + Qx + R = 0$, the discriminant is $D = Q^2 - 4PR$. If $D=0$, the equation has real and equal roots, given by $x = -\frac{Q}{2P}$.
3. Common Root: If two equations share a common root, that root must satisfy both equations.
4. Arithmetic Progression (A.P.): Three numbers $X, Y, Z$ are in A.P. if $Y - X = Z - Y$, which implies $2Y = X + Z$.

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Step-by-Step Solution:

Step 1: Analyze the first quadratic equation using the G.P. condition.

We are given that $a, b, c$ are three distinct numbers in G.P. This implies the relationship: $b^2 = ac \quad \text{(Equation 1)}$

Now, consider the first quadratic equation: $ax^2 + 2bx + c = 0$. Let's find the discriminant ($D$) of this equation. For a quadratic $Ax^2 + Bx + C = 0$, $D = B^2 - 4AC$. Here, $A=a$, $B=2b$, and $C=c$. $D = (2b)^2 - 4(a)(c)$ $D = 4b^2 - 4ac$ Substitute the G.P. condition $b^2 = ac$ from Equation 1 into the discriminant: $D = 4(ac) - 4ac$ $D = 0$ Explanation: A discriminant of zero ($D=0$) means the quadratic equation has real and equal roots. This is a significant simplification.

Since the roots are equal, let this common root be $\alpha$. The formula for equal roots is $x = -\frac{\text{coefficient of x}}{2 \times \text{coefficient of } x^2}$. $\alpha = -\frac{2b}{2a}$ $\alpha = -\frac{b}{a} \quad \text{(Equation 2)}$

Step 2: Utilize the common root condition for both equations.

We are given that the equations $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$ have a common root. From Step 1, we determined that the first equation has a unique root $\alpha = -b/a$. Explanation: Since $\alpha = -b/a$ is the only root of the first equation and it is also a common root with the second equation, it must satisfy the second equation.

Substitute $x = -b/a$ into the second equation $dx^2 + 2ex + f = 0$: $d\left(-\frac{b}{a}\right)^2 + 2e\left(-\frac{b}{a}\right) + f = 0$ $d\left(\frac{b^2}{a^2}\right) - \frac{2eb}{a} + f = 0$ To clear the denominators, multiply the entire equation by $a^2$: $d b^2 - 2eab + f a^2 = 0 \quad \text{(Equation 3)}$

Step 3: Simplify the equation using the G.P. condition.

We have Equation 3: $db^2 - 2eab + fa^2 = 0$. Recall from Equation 1 that $b^2 = ac$. Substitute this into Equation 3: $d(ac) - 2eab + fa^2 = 0$ $dac - 2eab + fa^2 = 0$ Explanation: This substitution is crucial because it introduces $c$ and allows us to simplify the expression further by factoring.

Since $a, b, c$ are distinct numbers in G.P., none of them can be zero. Specifically, $a \neq 0$ (otherwise, the first equation wouldn't be quadratic). Therefore, we can safely divide the entire equation by $a$: $\frac{dac}{a} - \frac{2eab}{a} + \frac{fa^2}{a} = \frac{0}{a}$ $dc - 2eb + fa = 0 \quad \text{(Equation 4)}$

Step 4: Manipulate the equation to match the given options.

Equation 4 is $dc - 2eb + fa = 0$. The options involve ratios like $\frac{d}{a}$, $\frac{e}{b}$, $\frac{f}{c}$. To obtain these ratios, we should divide Equation 4 by a term that will produce them. Dividing by $ac$ will achieve this, and since $a, c \neq 0$, $ac \neq 0$. Explanation: Dividing by $ac$ is a strategic move to transform the linear equation into a form that directly relates to the ratios presented in the options.

Divide Equation 4 by $ac$: $\frac{dc}{ac} - \frac{2eb}{ac} + \frac{fa}{ac} = \frac{0}{ac}$ $\frac{d}{a} - \frac{2eb}{ac} + \frac{f}{c} = 0$ Again, use the G.P. condition $b^2 = ac$ to substitute for $ac$ in the middle term: $\frac{d}{a} - \frac{2eb}{b^2} + \frac{f}{c} = 0$ Simplify the middle term: $\frac{d}{a} - \frac{2e}{b} + \frac{f}{c} = 0$ Rearrange the terms to isolate $\frac{2e}{b}$: $\frac{2e}{b} = \frac{d}{a} + \frac{f}{c}$

Step 5: Conclude the relationship.

The derived relationship $\frac{2e}{b} = \frac{d}{a} + \frac{f}{c}$ is the defining condition for three terms to be in Arithmetic Progression (A.P.). Let $X = \frac{d}{a}$, $Y = \frac{e}{b}$, and $Z = \frac{f}{c}$. The equation becomes $2Y = X + Z$. Explanation: This equation directly matches the A.P. condition, establishing the relationship between the ratios of the coefficients.

Therefore, the terms $\frac{d}{a}$, $\frac{e}{b}$, and $\frac{f}{c}$ are in A.P.

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Tips and Common Mistakes:

• Discriminant is Key: Always check the discriminant when coefficients are related. $D=0$ for the first equation was a critical simplification.
• Non-zero Coefficients: Ensure that any term you divide by is non-zero. In this problem, $a, b, c$ being distinct in G.P. implies they are all non-zero.
• Recognizing Patterns: Quickly identify the algebraic conditions for A.P. ($2Y=X+Z$) and G.P. ($B^2=AC$) to guide your manipulations.
• Strategic Substitution: Use the given G.P. condition ($b^2=ac$) at points where it simplifies the expression or helps achieve the desired form.

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Summary:

The problem leverages the properties of Geometric Progressions and Quadratic Equations. The fact that $a, b, c$ are in G.P. leads to the first quadratic equation having equal roots. This common root, $\alpha = -b/a$, is then substituted into the second quadratic equation. By further applying the G.P. condition ($b^2=ac$) and algebraic manipulation, we arrive at the relationship $\frac{2e}{b} = \frac{d}{a} + \frac{f}{c}$, which is the defining characteristic of an Arithmetic Progression. Thus, $\frac{d}{a}$, $\frac{e}{b}$, $\frac{f}{c}$ are in A.P.

The final answer is \boxed{(D)}.