If `x ne y` and the sequences `x, a_1, a_2, y and x, b_1, b_2, y` each are in A.P., then `((a_2 - a_1)/(b_2- b_1) )` is

Options

A. `2/3`
B. `3/2`
C. `1`
D. `3/4`

Answer

To solve the problem, we need to find the value of \(\frac{a_2 - a_1}{b_2 - b_1}\) given that the sequences \(x, a_1, a_2, y\) and \(x, b_1, b_2, y\) are in arithmetic progression (A.P.).

### Step-by-Step Solution:

1. **Understanding the A.P. Condition**:
Since \(x, a_1, a_2, y\) is in A.P., we can use the property of A.P. that states the middle terms are equal to the average of the outer terms. Therefore, we can write:
\[ a_1 + a_2 = \frac{x + y}{2} \quad \text{(1)} \]

2. **Finding \(a_2 - a_1\)**:
From equation (1), we can express \(a_2\) in terms of \(a_1\):
\[ a_2 = \frac{x + y}{2} - a_1 \]
To find \(a_2 - a_1\):
\[ a_2 - a_1 = \left(\frac{x + y}{2} - a_1\right) - a_1 = \frac{x + y}{2} - 2a_1 \]

3. **Applying the A.P. Condition to the Second Sequence**:
Similarly, for the sequence \(x, b_1, b_2, y\), we have:
\[ b_1 + b_2 = \frac{x + y}{2} \quad \text{(2)} \]

4. **Finding \(b_2 - b_1\)**:
From equation (2), we can express \(b_2\) in terms of \(b_1\):
\[ b_2 = \frac{x + y}{2} - b_1 \]
To find \(b_2 - b_1\):
\[ b_2 - b_1 = \left(\frac{x + y}{2} - b_1\right) - b_1 = \frac{x + y}{2} - 2b_1 \]

5. **Calculating \(\frac{a_2 - a_1}{b_2 - b_1}\)**:
Now we can substitute the expressions for \(a_2 - a_1\) and \(b_2 - b_1\) into our original expression:
\[ \frac{a_2 - a_1}{b_2 - b_1} = \frac{\frac{x + y}{2} - 2a_1}{\frac{x + y}{2} - 2b_1} \]

6. **Simplifying the Expression**:
Since both the numerator and denominator have the same structure, we can simplify:
\[ \frac{a_2 - a_1}{b_2 - b_1} = \frac{1}{1} = 1 \]

### Final Answer:
Thus, the value of \(\frac{a_2 - a_1}{b_2 - b_1}\) is \(1\).