If `a_1,a_2,a_3,a_4` are in `HP` , then `1/(a_1a_4)sum_(r=1)^3a_r a_(r+1)` is root of :

Options

A. `x^2 + 2x + 15 =0`
B. `x^2 + 2x -15 =0`
C. `x^2 - 2x +15 =0`
D. `x^2 - 2x -15=0`

Answer

To solve the problem, we need to show that if \( a_1, a_2, a_3, a_4 \) are in Harmonic Progression (HP), then the expression

\[ \frac{1}{a_1 a_4} \sum_{r=1}^{3} a_r a_{r+1} \]

is a root of a certain equation. Let's break down the solution step by step.

### Step 1: Understanding Harmonic Progression
If \( a_1, a_2, a_3, a_4 \) are in HP, it implies that the reciprocals \( \frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}, \frac{1}{a_4} \) are in Arithmetic Progression (AP).

### Step 2: Establishing the Common Difference
Let the common difference of the AP be \( d \). Then we can write the relationships as follows:

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

From these equations, we can express \( a_2, a_3, a_4 \) in terms of \( a_1 \) and \( d \).

### Step 3: Expressing Terms
From the first equation, we can express \( a_2 \):

\[ \frac{1}{a_2} = \frac{1}{a_1} + d \implies a_2 = \frac{a_1}{1 + a_1 d} \]

Similarly, we can find \( a_3 \) and \( a_4 \):

\[ \frac{1}{a_3} = \frac{1}{a_2} + d \implies a_3 = \frac{a_2}{1 + a_2 d} \]
\[ \frac{1}{a_4} = \frac{1}{a_3} + d \implies a_4 = \frac{a_3}{1 + a_3 d} \]

### Step 4: Calculating the Summation
Now we need to compute the summation \( \sum_{r=1}^{3} a_r a_{r+1} \):

\[ \sum_{r=1}^{3} a_r a_{r+1} = a_1 a_2 + a_2 a_3 + a_3 a_4 \]

### Step 5: Substituting Values
Substituting the values of \( a_2, a_3, a_4 \) into the summation and simplifying, we get:

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

### Step 6: Final Expression
Now we need to find the expression:

\[ \frac{1}{a_1 a_4} \left( a_1 a_2 + a_2 a_3 + a_3 a_4 \right) \]

After simplifying, we find that this expression equals \( 3 \).

### Step 7: Conclusion
Thus, we conclude that

\[ \frac{1}{a_1 a_4} \sum_{r=1}^{3} a_r a_{r+1} = 3 \]

This means that \( 3 \) is a root of the equation we are looking for.