Terms a, 1, b are in Arithmetic Progression and terms 1, a, b are in Geometric Progression Find a and b given `a ne b`

Options

A. A)2,4
B. B)`-2,1`
C. C)`4,1`
D. D)`-2,4`

Answer

To solve the problem where terms \( a, 1, b \) are in Arithmetic Progression (AP) and terms \( 1, a, b \) are in Geometric Progression (GP), we can follow these steps:

### Step 1: Set up the equations from the given conditions.

1. **For Arithmetic Progression (AP)**:
Since \( a, 1, b \) are in AP, we can use the property that the middle term is the average of the other two terms:
\[ 1 = \frac{a + b}{2} \]
Rearranging gives us:
\[ a + b = 2 \quad \text{(Equation 1)} \]

2. **For Geometric Progression (GP)**:
Since \( 1, a, b \) are in GP, we can use the property that the square of the middle term is equal to the product of the other two terms:
\[ a^2 = 1 \cdot b \quad \Rightarrow \quad a^2 = b \quad \text{(Equation 2)} \]

### Step 2: Substitute Equation 2 into Equation 1.

Now, we substitute the expression for \( b \) from Equation 2 into Equation 1:
\[ a + a^2 = 2 \]

### Step 3: Rearrange the equation.

Rearranging gives us:
\[ a^2 + a - 2 = 0 \]

### Step 4: Factor the quadratic equation.

Now we can factor the quadratic equation:
\[ (a - 1)(a + 2) = 0 \]
This gives us the solutions:
\[ a - 1 = 0 \quad \Rightarrow \quad a = 1 \]
\[ a + 2 = 0 \quad \Rightarrow \quad a = -2 \]

### Step 5: Find corresponding values of \( b \).

1. If \( a = 1 \):
\[ b = a^2 = 1^2 = 1 \]
However, this solution is not valid since \( a \neq b \).

2. If \( a = -2 \):
\[ b = a^2 = (-2)^2 = 4 \]

### Conclusion:

Thus, the values of \( a \) and \( b \) are:
\[ a = -2, \quad b = 4 \]

### Final Answer:
The solution is \( (a, b) = (-2, 4) \).

---