If a,b,c are in A.P. and if `(b-c) x^(2)+(c-a) x+(a-b)=0 and 2 (c+a) x^(2) +(b+c) x=0` have a common root, then

Options

A. `a^(2), b^(2), c^(2)` are in A.P.
B. `a,c,b` are in A.P.
C. `a^(2),c^(2),b^(2)` are in G.P.
D. none of these

Answer

To solve the problem, we need to analyze the two quadratic equations given and find the conditions under which they have a common root. We are given that \( a, b, c \) are in Arithmetic Progression (A.P.), which means:

\[ 2b = a + c \]

### Step 1: Write down the equations
The two quadratic equations are:

1. \( (b - c)x^2 + (c - a)x + (a - b) = 0 \) (Equation 1)
2. \( 2(c + a)x^2 + (b + c)x = 0 \) (Equation 2)

### Step 2: Factor the second equation
The second equation can be factored as follows:

\[ x(2(c + a)x + (b + c)) = 0 \]

This gives us one root as \( x = 0 \) and the other root from \( 2(c + a)x + (b + c) = 0 \):

\[ x = -\frac{b + c}{2(c + a)} \]

### Step 3: Check for common roots
Since we are looking for a common root, we need to check if \( x = 0 \) is a root of the first equation. Substituting \( x = 0 \) into Equation 1:

\[ (b - c)(0)^2 + (c - a)(0) + (a - b) = a - b \]

For \( x = 0 \) to be a root, we need:

\[ a - b = 0 \implies a = b \]

### Step 4: Substitute \( a = b \) into the A.P. condition
If \( a = b \), then substituting into the A.P. condition \( 2b = a + c \) gives:

\[ 2b = b + c \implies c = b \]

Thus, we have \( a = b = c \).

### Step 5: Check the conditions for A.P.
If \( a = b = c \), then all three numbers are equal, which trivially satisfies the condition of being in A.P.

### Step 6: Analyze the squares
Now, we need to check the squares:

- \( a^2 = b^2 = c^2 \)

### Conclusion
Since \( a^2, b^2, c^2 \) are all equal, they are in A.P. as well as in G.P. (Geometric Progression).

Thus, the final answer is that \( a^2, b^2, c^2 \) are in A.P. and G.P.