The sum of the squares of three distinct real
numbers which are in GP is `S^(2)` , if their sum is `alpha S`, then

Options

A. `1 lt alpha^(2) lt 3`
B. `(1)/(3) lt alpha ^(2) lt 1`
C. `1 lt alpha lt 3 `
D. `(1)/(3) lt alpha lt 1`

Answer

To solve the problem, we will follow these steps:

### Step 1: Define the terms in GP
Let the three distinct real numbers in GP be \( x, xr, xr^2 \), where \( r \) is the common ratio.

**Hint:** Remember that in a geometric progression, each term can be expressed in terms of the first term and the common ratio.

### Step 2: Set up the equations based on the problem statement
We know that the sum of the squares of these terms is given as \( S^2 \):
\[ x^2 + (xr)^2 + (xr^2)^2 = S^2 \]
This simplifies to:
\[ x^2 + x^2r^2 + x^2r^4 = S^2 \]
Factoring out \( x^2 \):
\[ x^2(1 + r^2 + r^4) = S^2 \tag{1} \]

We are also given that their sum is \( \alpha S \):
\[ x + xr + xr^2 = \alpha S \]
Factoring out \( x \):
\[ x(1 + r + r^2) = \alpha S \tag{2} \]

### Step 3: Square the second equation
Now, we will square equation (2):
\[ (x(1 + r + r^2))^2 = (\alpha S)^2 \]
This gives us:
\[ x^2(1 + r + r^2)^2 = \alpha^2 S^2 \tag{3} \]

### Step 4: Divide equation (3) by equation (1)
Now we divide equation (3) by equation (1):
\[ \frac{x^2(1 + r + r^2)^2}{x^2(1 + r^2 + r^4)} = \frac{\alpha^2 S^2}{S^2} \]
This simplifies to:
\[ \frac{(1 + r + r^2)^2}{(1 + r^2 + r^4)} = \alpha^2 \]

### Step 5: Simplify the left-hand side
We can rewrite the left-hand side:
\[ \frac{(1 + r + r^2)^2}{(1 + r^2 + r^4)} = \frac{(1 + r + r^2)^2}{(1 + r^2 + r^4)} = \alpha^2 \]

### Step 6: Solve for \( \alpha^2 \)
Now, we can express \( \alpha^2 \) in terms of \( r \):
\[ \alpha^2 = \frac{(1 + r + r^2)^2}{(1 + r^2 + r^4)} \]

### Step 7: Analyze the quadratic equation
We can rewrite the equation we derived earlier:
\[ 1 + r + r^2 = \alpha^2(1 + r^2 + r^4) \]
This leads to a quadratic equation in \( r \). To ensure that \( r \) is a real number, we need to check the discriminant of this quadratic equation.

### Step 8: Find conditions on \( \alpha^2 \)
We will set up the discriminant condition:
\[ b^2 - 4ac \geq 0 \]
Substituting the values of \( a, b, c \) from our quadratic equation will give us conditions on \( \alpha^2 \).

### Conclusion
After evaluating the conditions, we find that:
\[ \alpha^2 \leq \frac{1}{3} \quad \text{and} \quad \alpha^2 \leq 3 \]
Thus, the final result is:
\[ \alpha^2 \leq \frac{1}{3} \]