If a,b,c are in AP and `(a+2b-c)(2b+c-a)(c+a-b)=lambdaabc`, then `lambda` is

Options

A. 1
B. 2
C. 4
D. None of these

Explanation

To solve the problem, we will follow these steps:

1. **Understanding the condition of AP**: Since \(a\), \(b\), and \(c\) are in Arithmetic Progression (AP), we have the condition:
\[ b - a = c - b \]
This implies:
\[ 2b = a + c \]

2. **Substituting the AP condition**: We need to evaluate the expression:
\[ (a + 2b - c)(2b + c - a)(c + a - b) \]
Using the condition \(2b = a + c\), we can simplify each term in the product.

- For the first term:
\[ a + 2b - c = a + (a + c) - c = 2a \]

- For the second term:
\[ 2b + c - a = (a + c) + c - a = 2c \]

- For the third term:
\[ c + a - b = c + a - \frac{a + c}{2} = c + a - \frac{a}{2} - \frac{c}{2} = \frac{2c + 2a - a - c}{2} = \frac{c + a}{2} = \frac{a + c}{2} \]

3. **Putting it all together**: Now substituting these simplified terms back into the expression:
\[ (2a)(2c)\left(\frac{a + c}{2}\right) \]
This simplifies to:
\[ 2a \cdot 2c \cdot \frac{a + c}{2} = 2ac(a + c) \]

4. **Setting the expression equal to \(\lambda abc\)**: We have:
\[ 2ac(a + c) = \lambda abc \]
Dividing both sides by \(abc\) (assuming \(a\), \(b\), and \(c\) are non-zero):
\[ \lambda = \frac{2ac(a + c)}{abc} = \frac{2(a + c)}{b} \]

5. **Using the AP condition again**: Since \(2b = a + c\), we can substitute \(a + c\) with \(2b\):
\[ \lambda = \frac{2(2b)}{b} = 4 \]

Thus, the value of \(\lambda\) is:
\[ \lambda = 4 \]