The sum of the first three terms of a G.P is S and their product is 27 . Then all such S lie in :

Options

A. ` (-oo ,9]`
B. `[-3,oo)`
C. `(- oo , -9 ] uu [ ,oo)`
D. `(-oo , -3 ]uu [ 9 ,oo)`

Answer

To solve the problem, we need to find the range of the sum \( S \) of the first three terms of a geometric progression (G.P.) given that their product is 27.

### Step-by-Step Solution:

1. **Define the terms of the G.P.**:
Let the three terms of the G.P. be \( a/r, a, ar \), where \( a \) is the middle term and \( r \) is the common ratio.

2. **Set up the product equation**:
The product of these terms is given by:
\[ \frac{a}{r} \cdot a \cdot ar = 27 \]
Simplifying this, we get:
\[ a^3 = 27 \]
Therefore, we find:
\[ a = 3 \]

3. **Set up the sum equation**:
The sum of the first three terms is:
\[ \frac{a}{r} + a + ar = S \]
Substituting \( a = 3 \):
\[ \frac{3}{r} + 3 + 3r = S \]
This simplifies to:
\[ S = 3 \left( \frac{1}{r} + 1 + r \right) \]

4. **Introduce a new variable**:
Let \( x = r + \frac{1}{r} \). Then we can rewrite \( S \) as:
\[ S = 3(x + 1) \]

5. **Analyze the variable \( x \)**:
The expression \( x = r + \frac{1}{r} \) has a minimum value of 2 for \( r > 0 \) (by the AM-GM inequality). For \( r < 0 \), \( x \) can take values less than or equal to -2.

6. **Determine the range of \( S \)**:
- For \( r > 0 \):
\[ x \geq 2 \implies S \geq 3(2 + 1) = 9 \]
- For \( r < 0 \):
\[ x \leq -2 \implies S \leq 3(-2 + 1) = -3 \]

7. **Combine the results**:
Thus, we find that:
\[ S \geq 9 \quad \text{or} \quad S \leq -3 \]
This can be expressed in interval notation as:
\[ S \in (-\infty, -3] \cup [9, \infty) \]

### Final Answer:
The values of \( S \) lie in the intervals:
\[ S \in (-\infty, -3] \cup [9, \infty) \]