If `(a_2a_3)/(a_1a_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. A.P.
B. G.P
C. H.P.
D. None of these

Answer

To solve the problem, we need to analyze the given equations:

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 this common value as \(k\). Thus, we can write:

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

### Step 1: Equating the first two equations

From the first two equations, we have:

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

Cross-multiplying gives:

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

Expanding both sides:

\[ a_2 a_3 a_1 + a_2 a_3 a_4 = a_2 a_1 a_4 + a_3 a_1 a_4 \]

Rearranging terms:

\[ a_2 a_3 a_1 + a_2 a_3 a_4 - a_2 a_1 a_4 - a_3 a_1 a_4 = 0 \]

### Step 2: Rearranging the equation

Rearranging gives:

\[ a_2 a_3 a_1 + a_2 a_3 a_4 = a_2 a_1 a_4 + a_3 a_1 a_4 \]

Factoring out common terms:

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

### Step 3: Equating the second and third equations

Now, let's take the second and third equations:

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

Cross-multiplying gives:

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

Expanding both sides:

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

### Step 4: Rearranging and simplifying

Rearranging gives:

\[ a_2 a_1 - a_2 a_4 + a_3 a_1 - a_3 a_4 - 3(a_2 a_1 + a_2 a_4 - a_3 a_1 - a_3 a_4) = 0 \]

This can be simplified to find a relationship between \(a_1, a_2, a_3, a_4\).

### Step 5: Conclusion

From the derived equations, we can see that the terms \( \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \frac{1}{a_4} \) are in Arithmetic Progression (AP). Therefore, the original terms \( a_1, a_2, a_3, a_4 \) are in Harmonic Progression (HP).

Thus, the answer is that \( a_1, a_2, a_3, a_4 \) are in **Harmonic Progression (HP)**.