If `a ,b ,c` are three distinct real numbers in G.P. and `a+b+c=x b ,` then prove that either `x<<-1orx>>3.`

Options

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

Answer

To solve the problem, we start with the given conditions:

1. \( a, b, c \) are three distinct real numbers in a geometric progression (G.P.).
2. The relationship \( a + b + c = x b \) holds.

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

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

### Step 3: Factor out \( a \)
Factoring \( a \) out from the left side:
\[ a(1 + r + r^2) = x(ar) \]

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

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

### Step 6: Identify the quadratic form
This is a quadratic equation in \( r \):
\[ r^2 + (1 - x)r + 1 = 0 \]

### Step 7: Determine the discriminant
For \( r \) to be real, 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 8: Solve the inequality
Expanding the discriminant:
\[ (1 - x)^2 > 4 \]
Taking square roots gives us two cases:
1. \( 1 - x > 2 \) or
2. \( 1 - x < -2 \)

#### Case 1:
\[ 1 - x > 2 \implies -x > 1 \implies x < -1 \]

#### Case 2:
\[ 1 - x < -2 \implies -x < -3 \implies x > 3 \]

### Conclusion
Thus, we have proven that either:
\[ x < -1 \quad \text{or} \quad x > 3 \]