If a, b and c be three distinct real number in G.P. and `a + b + c = xb`, then x cannot be

Options

A. 2
B. 4
C. -3
D. -2

Explanation

To solve the problem, we start with the given conditions that \( a, b, c \) are three distinct real numbers in geometric progression (G.P.) and that \( a + b + c = xb \).

### Step 1: Express the terms in G.P.
Since \( a, b, c \) are in G.P., we can express them as:
- \( a = a \)
- \( b = ar \)
- \( c = ar^2 \)

where \( r \) is the common ratio.

### Step 2: Substitute into the equation
Substituting \( b \) and \( c \) into the equation \( a + b + c = xb \):
\[ a + ar + ar^2 = xb \]
This simplifies to:
\[ a + ar + ar^2 = x(ar) \]

### Step 3: Factor out common terms
We can factor out \( a \) from the left side:
\[ a(1 + r + r^2) = xar \]

### Step 4: Cancel \( a \) (assuming \( a \neq 0 \))
Since \( a \) is non-zero (as \( a, b, c \) are distinct), we can divide both sides by \( a \):
\[ 1 + r + r^2 = xr \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ r^2 + r + (1 - xr) = 0 \]
This can be rewritten as:
\[ r^2 + (1 - x)r + 1 = 0 \]

### Step 6: Use the discriminant condition
For \( r \) to have real solutions, the discriminant must be greater than zero:
\[ D = (1 - x)^2 - 4 \cdot 1 \cdot 1 > 0 \]
This simplifies to:
\[ (1 - x)^2 - 4 > 0 \]

### Step 7: Solve the inequality
Expanding the inequality:
\[ (1 - x)^2 > 4 \]
Taking square roots:
\[ |1 - x| > 2 \]
This leads to two cases:
1. \( 1 - x > 2 \) or
2. \( 1 - x < -2 \)

From the first case:
\[ 1 - x > 2 \implies -x > 1 \implies x < -1 \]

From the second case:
\[ 1 - x < -2 \implies -x < -3 \implies x > 3 \]

### Step 8: Combine the results
Thus, we find that \( x \) cannot be in the interval:
\[ x \in (-1, 3) \]

### Conclusion
The values that \( x \) cannot take are:
\[ x \text{ cannot be } -1 \text{ or } 3. \]