Let `alpha, beta, gamma` be three numbers such that `(1)/(alpha)+(1)/(beta)+(1)/(gamma)=(1)/(2), (1)/(alpha^(2))+(1)/(beta^(2))+(1)/(gamma^(2))=(9)/(4) and alpha + beta + gamma=2`, then
`{:("Column-I","Column-II"),("(a) " alpha beta gamma ,"(p) " 6),("(b) "Sigma alpha beta, "(q) "8),("(c) "Sigma alpha^(2) , "(r) "-2), ("(d "Sigma alpha^(3),"(s) "-1):}`

Answer

To solve the problem, we are given three conditions involving the numbers \( \alpha, \beta, \gamma \):

1. \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{1}{2} \)
2. \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{9}{4} \)
3. \( \alpha + \beta + \gamma = 2 \)

We need to find the values of \( \alpha \beta \gamma \), \( \sigma \alpha \beta \) (which is \( \alpha \beta + \beta \gamma + \gamma \alpha \)), \( \sigma \alpha^2 \) (which is \( \alpha^2 + \beta^2 + \gamma^2 \)), and \( \sigma \alpha^3 \) (which is \( \alpha^3 + \beta^3 + \gamma^3 \)).

### Step 1: Find \( \alpha \beta \gamma \)

From the first condition, we can rewrite it as:

\[ \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\beta \gamma + \alpha \gamma + \alpha \beta}{\alpha \beta \gamma} = \frac{1}{2} \]

This implies:

\[ \beta \gamma + \alpha \gamma + \alpha \beta = \frac{1}{2} \alpha \beta \gamma \quad \text{(1)} \]

### Step 2: Find \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} \)

Using the identity:

\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} = \frac{(\beta \gamma + \alpha \gamma + \alpha \beta)^2 - 2\alpha \beta \gamma(\alpha + \beta + \gamma)}{(\alpha \beta \gamma)^2} \]

Substituting the known values:

\[ \frac{9}{4} = \frac{(\frac{1}{2} \alpha \beta \gamma)^2 - 2\alpha \beta \gamma \cdot 2}{(\alpha \beta \gamma)^2} \]

This leads to:

\[ 9(\alpha \beta \gamma)^2 = 4 \left( \frac{1}{4} (\alpha \beta \gamma)^2 - 4 \alpha \beta \gamma \right) \]

### Step 3: Solve for \( \alpha \beta \gamma \)

From the above equation, we can simplify and solve for \( \alpha \beta \gamma \):

After simplification, we find:

\[ \alpha \beta \gamma = -2 \quad \text{(2)} \]

### Step 4: Find \( \sigma \alpha \beta \)

Using the relation:

\[ \sigma \alpha = \alpha + \beta + \gamma = 2 \]

We can substitute this into the equation from step 1 to find \( \sigma \alpha \beta \):

\[ \sigma \alpha \beta = \frac{1}{2} \cdot (-2) = -1 \quad \text{(3)} \]

### Step 5: Find \( \sigma \alpha^2 \)

Using the identity:

\[ \sigma \alpha^2 = (\alpha + \beta + \gamma)^2 - 2\sigma \alpha \beta \]

Substituting the known values:

\[ \sigma \alpha^2 = 2^2 - 2(-1) = 4 + 2 = 6 \quad \text{(4)} \]

### Step 6: Find \( \sigma \alpha^3 \)

Using the identity:

\[ \sigma \alpha^3 = (\alpha + \beta + \gamma)(\sigma \alpha^2) - \sigma \alpha \beta \cdot \sigma \alpha \]

Substituting the known values:

\[ \sigma \alpha^3 = 2 \cdot 6 - (-1)(2) = 12 + 2 = 14 \quad \text{(5)} \]

### Final Results

Now we have:

- \( \alpha \beta \gamma = -2 \)
- \( \sigma \alpha \beta = -1 \)
- \( \sigma \alpha^2 = 6 \)
- \( \sigma \alpha^3 = 14 \)

### Summary of Results

- \( \alpha \beta \gamma = -2 \) (matches option \( r \))
- \( \sigma \alpha \beta = -1 \) (matches option \( s \))
- \( \sigma \alpha^2 = 6 \) (matches option \( p \))
- \( \sigma \alpha^3 = 14 \) (not in options)