If a, b, c are positive real numbers which are in G. P., then the equation $ax^{2}+2bx+c=0$ and $dx^{2}+2ex+f=0$ have a common root, if $\frac{d}{a},\frac{e}{b},\frac{f}{c}$ are in:

Question

Accepted Answer

To determine why Option 1 is correct, we need to analyze the conditions under which the given quadratic equations have a common root, given that $a, b, c$ are in geometric progression (G.P.) and $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in arithmetic progression (A.P.).

### Explanation:

1. **Understanding the Problem:**
   - We have two quadratic equations: $ax^2 + 2bx + c = 0$ and $dx^2 + 2ex + f = 0$.
   - These equations have a common root.
   - $a, b, c$ are in G.P., meaning $b^2 = ac$.
   - We need to show that $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.

2. **Common Root Condition:**
   - Let the common root be $\alpha$.
   - Substituting $\alpha$ into both equations, we get:
     $$
     a\alpha^2 + 2b\alpha + c = 0
     $$
     $$
     d\alpha^2 + 2e\alpha + f = 0
     $$
   - These can be rewritten as:
     $$
     a\alpha^2 + 2b\alpha + c = 0 \quad \text{(1)}
     $$
     $$
     d\alpha^2 + 2e\alpha + f = 0 \quad \text{(2)}
     $$

3. **Using the Condition of A.P.:**
   - For $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ to be in A.P., the difference between consecutive terms must be constant:
     $$
     \frac{e}{b} - \frac{d}{a} = \frac{f}{c} - \frac{e}{b}
     $$
   - Simplifying, we get:
     $$
     \frac{e}{b} - \frac{d}{a} = \frac{f}{c} - \frac{e}{b}
     $$
     $$
     2\left(\frac{e}{b}\right) = \frac{d}{a} + \frac{f}{c}
     $$

4. **Why Option 1 is Correct:**
   - The condition derived above is precisely the condition for the terms to be in A.P. This means that the sequence $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ satisfies the property of an arithmetic progression.
   - Therefore, Option 1 is correct because the condition for a common root aligns with the terms being in A.P.

5. **Why Other Options are Incorrect:**
   - **G.P. (Option 2):** For terms to be in G.P., the ratio of consecutive terms must be constant. The derived condition does not satisfy this requirement.
   - **H.P. (Option 3):** For terms to be in H.P., the reciprocals of the terms must be in A.P. The derived condition does not satisfy this requirement either.
   - **None of the three (Option 4):** Since we have shown that the terms are in A.P., this option is incorrect.

In conclusion, the correct answer is Option 1, as the condition for a common root leads to the terms $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ being in arithmetic progression.