Suppose a, b and c are in Arithmetic Progression and `a^2, b^2, and c^2` are in Geometric Progression. If a < b < c and `a+ b+c = 3/2` then the value of a =

Options

A. `(1)/(2sqrt2)`
B. `(1)/(2sqrt3)`
C. `1/2 - (1)/(sqrt3)`
D. `1/2 - (1)/(sqrt2)`

Answer

To solve the problem step by step, let's break it down:

### Step 1: Understand the conditions
We know that \( a, b, c \) are in Arithmetic Progression (AP) and \( a^2, b^2, c^2 \) are in Geometric Progression (GP). We also have the condition \( a + b + c = \frac{3}{2} \).

### Step 2: Express \( b \) and \( c \) in terms of \( a \)
Since \( a, b, c \) are in AP, we can express them as:
- Let \( b = x \) (the middle term)
- Let \( a = x - d \) (the first term)
- Let \( c = x + d \) (the third term)

### Step 3: Write the sum equation
From the condition \( a + b + c = \frac{3}{2} \):
\[ (x - d) + x + (x + d) = \frac{3}{2} \]
This simplifies to:
\[ 3x = \frac{3}{2} \]

### Step 4: Solve for \( x \)
Dividing both sides by 3 gives:
\[ x = \frac{1}{2} \]

### Step 5: Substitute \( x \) back to find \( a \) and \( c \)
Now substituting \( x \) back into the expressions for \( a \) and \( c \):
- \( a = x - d = \frac{1}{2} - d \)
- \( c = x + d = \frac{1}{2} + d \)

### Step 6: Use the GP condition
The condition that \( a^2, b^2, c^2 \) are in GP means:
\[ b^2 = \sqrt{a^2 \cdot c^2} \]
Substituting the values:
\[ \left(\frac{1}{2}\right)^2 = \sqrt{(\frac{1}{2} - d)^2 \cdot (\frac{1}{2} + d)^2} \]
Squaring both sides gives:
\[ \frac{1}{4} = (\frac{1}{2} - d)(\frac{1}{2} + d) \]
This simplifies to:
\[ \frac{1}{4} = \frac{1}{4} - d^2 \]

### Step 7: Solve for \( d^2 \)
Rearranging gives:
\[ d^2 = 0 \]
This implies \( d = 0 \).

### Step 8: Find the value of \( a \)
Now substituting \( d = 0 \) back into the expression for \( a \):
\[ a = \frac{1}{2} - 0 = \frac{1}{2} \]

### Conclusion
Thus, the value of \( a \) is:
\[ \boxed{\frac{1}{2}} \]