If roots of `2x^(2) - 3x + p ` are `1/alpha, 1/beta` and that of `2x^(2) - 7x +q = 0` are `1/gamma, 1/delta` are such that `alpha, beta, gamma, delta` are in HP then p+q =

Options

A. 5
B. 7
C. 11
D. 13

Explanation

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Understanding the Roots
We are given two quadratic equations:
1. \(2x^2 - 3x + p = 0\) with roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \)
2. \(2x^2 - 7x + q = 0\) with roots \( \frac{1}{\gamma} \) and \( \frac{1}{\delta} \)

Since \( \alpha, \beta, \gamma, \delta \) are in Harmonic Progression (HP), their reciprocals \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta} \) will be in Arithmetic Progression (AP).

### Step 2: Sum of Roots
For the first equation:
- The sum of the roots \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{2} \)

For the second equation:
- The sum of the roots \( \frac{1}{\gamma} + \frac{1}{\delta} = \frac{7}{2} \)

### Step 3: Expressing in Terms of \( \alpha \) and \( d \)
Since \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta} \) are in AP, we can express them as:
- \( \frac{1}{\beta} = \frac{1}{\alpha} + d \)
- \( \frac{1}{\gamma} = \frac{1}{\alpha} + 2d \)
- \( \frac{1}{\delta} = \frac{1}{\alpha} + 3d \)

### Step 4: Setting Up Equations
From the sum of roots for the first equation:
\[ \frac{1}{\alpha} + \left(\frac{1}{\alpha} + d\right) = \frac{3}{2} \]
This simplifies to:
\[ \frac{2}{\alpha} + d = \frac{3}{2} \quad \text{(1)} \]

From the sum of roots for the second equation:
\[ \left(\frac{1}{\alpha} + 2d\right) + \left(\frac{1}{\alpha} + 3d\right) = \frac{7}{2} \]
This simplifies to:
\[ \frac{2}{\alpha} + 5d = \frac{7}{2} \quad \text{(2)} \]

### Step 5: Solving the Equations
Now we will solve equations (1) and (2) together.

From equation (1):
\[ d = \frac{3}{2} - \frac{2}{\alpha} \]

Substituting \( d \) in equation (2):
\[ \frac{2}{\alpha} + 5\left(\frac{3}{2} - \frac{2}{\alpha}\right) = \frac{7}{2} \]
This simplifies to:
\[ \frac{2}{\alpha} + \frac{15}{2} - \frac{10}{\alpha} = \frac{7}{2} \]
Combining like terms:
\[ -\frac{8}{\alpha} + \frac{15}{2} = \frac{7}{2} \]
Subtracting \( \frac{15}{2} \) from both sides:
\[ -\frac{8}{\alpha} = \frac{7}{2} - \frac{15}{2} = -4 \]
Thus:
\[ \frac{8}{\alpha} = 4 \implies \alpha = 2 \]

### Step 6: Finding \( p \) and \( q \)
Now we can find \( p \) and \( q \).

For \( p \):
Using the product of roots:
\[ \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{p}{2} \]
We know \( \frac{1}{\alpha} = \frac{1}{2} \) and \( \frac{1}{\beta} = \frac{1}{\alpha} + d = \frac{1}{2} + \frac{1}{2} = 1 \):
\[ \frac{1}{2} \cdot 1 = \frac{p}{2} \implies \frac{1}{2} = \frac{p}{2} \implies p = 1 \]

For \( q \):
Using the product of roots:
\[ \frac{1}{\gamma} \cdot \frac{1}{\delta} = \frac{q}{2} \]
We have:
\[ \frac{1}{\gamma} = \frac{1}{\alpha} + 2d = \frac{1}{2} + 1 = \frac{3}{2} \]
\[ \frac{1}{\delta} = \frac{1}{\alpha} + 3d = \frac{1}{2} + \frac{3}{2} = 2 \]
Thus:
\[ \frac{3}{2} \cdot 2 = \frac{q}{2} \implies 3 = \frac{q}{2} \implies q = 6 \]

### Step 7: Finding \( p + q \)
Finally, we calculate:
\[ p + q = 1 + 6 = 7 \]

### Final Answer
\[ \boxed{7} \]