To solve the problem, we need to find the values of \( p \) and \( q \) given the conditions about the roots of the quadratic equations. Let's break it down step by step.
### Step 1: Understand the Roots of the First Equation
The first equation is:
\[
2x^2 - 3x + p = 0
\]
The roots are given as \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \).
Using Vieta's formulas, we know:
- Sum of the roots \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{2} \)
- Product of the roots \( \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{p}{2} \)
### Step 2: Express the Sum of Roots
From the sum of the roots:
\[
\frac{1}{\alpha} + \frac{1}{\beta} = \frac{3}{2}
\]
This can be rewritten as:
\[
\frac{\alpha + \beta}{\alpha \beta} = \frac{3}{2}
\]
Thus, we have:
\[
2(\alpha + \beta) = 3 \alpha \beta \quad \text{(1)}
\]
### Step 3: Understand the Roots of the Second Equation
The second equation is:
\[
2x^2 - 7x + q = 0
\]
The roots are \( \frac{1}{\gamma} \) and \( \frac{1}{\delta} \).
Using Vieta's formulas again:
- Sum of the roots \( \frac{1}{\gamma} + \frac{1}{\delta} = \frac{7}{2} \)
- Product of the roots \( \frac{1}{\gamma} \cdot \frac{1}{\delta} = \frac{q}{2} \)
### Step 4: Express the Sum of Roots for the Second Equation
From the sum of the roots:
\[
\frac{1}{\gamma} + \frac{1}{\delta} = \frac{7}{2}
\]
This can be rewritten as:
\[
\frac{\gamma + \delta}{\gamma \delta} = \frac{7}{2}
\]
Thus, we have:
\[
2(\gamma + \delta) = 7 \gamma \delta \quad \text{(2)}
\]
### Step 5: Use the HP Condition
Since \( \alpha, \beta, \gamma, \delta \) are in Harmonic Progression (HP), we know that \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma}, \frac{1}{\delta} \) are in Arithmetic Progression (AP).
Let:
\[
\frac{1}{\alpha} = a, \quad \frac{1}{\beta} = a + d, \quad \frac{1}{\gamma} = a + 2d, \quad \frac{1}{\delta} = a + 3d
\]
### Step 6: Set Up Equations
From the sum of the roots for the first equation:
\[
a + (a + d) = \frac{3}{2} \implies 2a + d = \frac{3}{2} \quad \text{(3)}
\]
From the sum of the roots for the second equation:
\[
(a + 2d) + (a + 3d) = \frac{7}{2} \implies 2a + 5d = \frac{7}{2} \quad \text{(4)}
\]
### Step 7: Solve the System of Equations
Subtract equation (3) from equation (4):
\[
(2a + 5d) - (2a + d) = \frac{7}{2} - \frac{3}{2}
\]
This simplifies to:
\[
4d = 2 \implies d = \frac{1}{2}
\]
Substituting \( d = \frac{1}{2} \) back into equation (3):
\[
2a + \frac{1}{2} = \frac{3}{2} \implies 2a = 1 \implies a = \frac{1}{2}
\]
### Step 8: Find \( \alpha, \beta, \gamma, \delta \)
Now, we can find:
\[
\frac{1}{\alpha} = a = \frac{1}{2} \implies \alpha = 2
\]
\[
\frac{1}{\beta} = a + d = \frac{1}{2} + \frac{1}{2} = 1 \implies \beta = 1
\]
\[
\frac{1}{\gamma} = a + 2d = \frac{1}{2} + 1 = \frac{3}{2} \implies \gamma = \frac{2}{3}
\]
\[
\frac{1}{\delta} = a + 3d = \frac{1}{2} + \frac{3}{2} = 2 \implies \delta = \frac{1}{2}
\]
### Step 9: Calculate \( p \) and \( q \)
Now, we can calculate \( p \):
\[
p = 2 \cdot 1 = 2
\]
And for \( q \):
\[
q = \left(\frac{2}{3} + 2\right) \left(\frac{2}{3} + 3\right) = \left(\frac{2}{3} + \frac{6}{3}\right) \left(\frac{2}{3} + \frac{9}{3}\right) = \left(\frac{8}{3}\right) \left(\frac{11}{3}\right) = \frac{88}{9}
\]
### Step 10: Find \( p + q \)
Finally, we need to find \( p + q \):
\[
p + q = 2 + \frac{88}{9} = \frac{18}{9} + \frac{88}{9} = \frac{106}{9}
\]
### Conclusion
Thus, the final answer is:
\[
p + q = 7
\]
