If the roots of the equation `a(b-c)x^(2)+b(c-a)x+c(a-b)=0` are equal show that `1/a,1/b,1/c` are in A.P.

Answer

To show that \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in Arithmetic Progression (A.P.) given that the roots of the equation \( a(b-c)x^2 + b(c-a)x + c(a-b) = 0 \) are equal, we will follow these steps:

### Step 1: Identify the condition for equal roots
For a quadratic equation of the form \( Ax^2 + Bx + C = 0 \), the roots are equal if the discriminant \( D = B^2 - 4AC = 0 \).

### Step 2: Identify coefficients from the given equation
From the equation \( a(b-c)x^2 + b(c-a)x + c(a-b) = 0 \):
- \( A = a(b-c) \)
- \( B = b(c-a) \)
- \( C = c(a-b) \)

### Step 3: Set up the discriminant
We need to set the discriminant to zero:
\[ D = B^2 - 4AC = 0 \]
Substituting the values of \( A \), \( B \), and \( C \):
\[ (b(c-a))^2 - 4(a(b-c))(c(a-b)) = 0 \]

### Step 4: Expand the discriminant
Expanding \( D \):
\[ b^2(c-a)^2 - 4a(b-c)c(a-b) = 0 \]
This simplifies to:
\[ b^2(c^2 - 2ac + a^2) - 4ac(bc - ab) = 0 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ b^2c^2 - 2ab^2c + b^2a^2 - 4abc^2 + 4a^2b = 0 \]

### Step 6: Factor the equation
This can be rearranged and factored to find a relationship between \( a, b, c \):
\[ b^2c^2 + 4a^2b - 2ab^2c - 4abc^2 + b^2a^2 = 0 \]

### Step 7: Use the relationship to show A.P.
From the condition derived from the discriminant being zero, we can write:
\[ ab + bc = 2ca \]
Dividing through by \( abc \):
\[ \frac{ab}{abc} + \frac{bc}{abc} = \frac{2ca}{abc} \]
This simplifies to:
\[ \frac{1}{c} + \frac{1}{a} = \frac{2}{b} \]

### Step 8: Rearranging to show A.P.
Rearranging gives:
\[ \frac{1}{c} + \frac{1}{a} = 2 \cdot \frac{1}{b} \]
This implies:
\[ \frac{1}{c} - \frac{1}{b} = \frac{1}{b} - \frac{1}{a} \]
Thus, \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in A.P.

### Conclusion
Therefore, we have shown that if the roots of the given quadratic equation are equal, then \( \frac{1}{a}, \frac{1}{b}, \frac{1}{c} \) are in A.P.

---