If the roots of the equation `4x^3 -12x^2 +11x +k=0` are in arithmetic progression then k=

Options

A. `-3`
B. `1`
C. `2`
D. `3`

Answer

To find the value of \( k \) in the equation \( 4x^3 - 12x^2 + 11x + k = 0 \) given that the roots are in arithmetic progression, we can follow these steps:

### Step 1: Define the Roots
Let the roots of the cubic equation be \( \alpha - d, \alpha, \alpha + d \), where \( \alpha \) is the middle root and \( d \) is the common difference.

### Step 2: Use the Sum of Roots
According to Vieta's formulas, the sum of the roots can be expressed as:
\[ (\alpha - d) + \alpha + (\alpha + d) = -\frac{b}{a} \]
Here, \( b = -12 \) and \( a = 4 \). Thus:
\[ 3\alpha = -\frac{-12}{4} = 3 \]
From this, we can solve for \( \alpha \):
\[ \alpha = 1 \]

### Step 3: Use the Product of Roots
Next, we can use the product of the roots, which is given by:
\[ (\alpha - d) \cdot \alpha \cdot (\alpha + d) = -\frac{k}{a} \]
Substituting \( a = 4 \) and \( \alpha = 1 \):
\[ (1 - d) \cdot 1 \cdot (1 + d) = -\frac{k}{4} \]
This simplifies to:
\[ 1^2 - d^2 = -\frac{k}{4} \]
So:
\[ 1 - d^2 = -\frac{k}{4} \]

### Step 4: Use the Sum of the Products of Roots
We also have the sum of the products of the roots taken two at a time:
\[ (\alpha - d)\alpha + \alpha(\alpha + d) + (\alpha - d)(\alpha + d) = \frac{c}{a} \]
Where \( c = 11 \) and \( a = 4 \):
\[ (1 - d) \cdot 1 + 1 \cdot (1 + d) + (1 - d)(1 + d) = \frac{11}{4} \]
This simplifies to:
\[ (1 - d) + (1 + d) + (1^2 - d^2) = \frac{11}{4} \]
Combining terms gives:
\[ 2 + 1 - d^2 = \frac{11}{4} \]
Thus:
\[ 3 - d^2 = \frac{11}{4} \]

### Step 5: Solve for \( d^2 \)
Now we can solve for \( d^2 \):
\[ 3 - \frac{11}{4} = d^2 \]
Converting 3 to a fraction:
\[ \frac{12}{4} - \frac{11}{4} = d^2 \]
So:
\[ d^2 = \frac{1}{4} \]

### Step 6: Substitute \( d^2 \) Back
Now substitute \( d^2 \) back into the equation for \( k \):
\[ 1 - \frac{1}{4} = -\frac{k}{4} \]
This simplifies to:
\[ \frac{3}{4} = -\frac{k}{4} \]
Multiplying both sides by -4 gives:
\[ k = -3 \]

### Conclusion
Thus, the value of \( k \) is:
\[ \boxed{-3} \]