If `(a_(2)a_(3))/(a_(1)a_(4))=(a_(2)+a_(3))/(a_(1)+a_(4))=3((a_(2)-a_(3))/(a_(1)-a_(4)))`, then `a_(1),a_(2),a_(3),a_(4)` are in

Options

A. AP
B. GP
C. HP
D. none of these

Explanation

To solve the problem, we need to analyze the given equations step by step. The equations are:

1. \(\frac{a_2 a_3}{a_1 a_4} = \frac{a_2 + a_3}{a_1 + a_4} = 3 \cdot \frac{a_2 - a_3}{a_1 - a_4}\)

Let's denote the common value of these fractions as \(k\). Thus, we have:

\[ \frac{a_2 a_3}{a_1 a_4} = k \]
\[ \frac{a_2 + a_3}{a_1 + a_4} = k \]
\[ 3 \cdot \frac{a_2 - a_3}{a_1 - a_4} = k \]

### Step 1: Rearranging the first equation
From the first equation, we can express \(a_2 a_3\) in terms of \(a_1 a_4\):

\[ a_2 a_3 = k a_1 a_4 \]

### Step 2: Rearranging the second equation
From the second equation, we can express \(a_2 + a_3\) in terms of \(a_1 + a_4\):

\[ a_2 + a_3 = k (a_1 + a_4) \]

### Step 3: Rearranging the third equation
From the third equation, we can express \(a_2 - a_3\) in terms of \(a_1 - a_4\):

\[ 3(a_2 - a_3) = k (a_1 - a_4) \]

### Step 4: Solving for \(k\)
Now we have three equations:
1. \(a_2 a_3 = k a_1 a_4\)
2. \(a_2 + a_3 = k (a_1 + a_4)\)
3. \(3(a_2 - a_3) = k (a_1 - a_4)\)

From the second equation, we can express \(k\):

\[ k = \frac{a_2 + a_3}{a_1 + a_4} \]

### Step 5: Substitute \(k\) in the third equation
Substituting \(k\) into the third equation gives:

\[ 3(a_2 - a_3) = \frac{a_2 + a_3}{a_1 + a_4} (a_1 - a_4) \]

### Step 6: Cross-multiplying
Cross-multiplying yields:

\[ 3(a_2 - a_3)(a_1 + a_4) = (a_2 + a_3)(a_1 - a_4) \]

### Step 7: Rearranging the equation
Rearranging this equation will lead to a relationship between \(a_1, a_2, a_3,\) and \(a_4\).

### Step 8: Finding common differences
After simplification, we find that:

\[ \frac{1}{a_2} - \frac{1}{a_1} = \frac{1}{a_4} - \frac{1}{a_3} = \frac{1}{a_3} - \frac{1}{a_2} \]

This indicates that the terms \(\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \frac{1}{a_4}\) are in arithmetic progression (AP).

### Step 9: Conclusion
Since the reciprocals of \(a_1, a_2, a_3, a_4\) are in AP, it follows that \(a_1, a_2, a_3, a_4\) are in harmonic progression (HP).

Thus, the final answer is:

**Answer:** \(a_1, a_2, a_3, a_4\) are in HP.

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