IF the roots of `x^3 - 13x^2 +kx - 27=0` are in G.P then k=

Options

A. `-30`
B. `30`
C. `39`
D. `-39`

Answer

To solve the problem, we need to find the value of \( k \) given that the roots of the polynomial \( x^3 - 13x^2 + kx - 27 = 0 \) are in geometric progression (G.P.).

### Step-by-Step Solution:

1. **Identify the Roots in G.P.**:
Let the roots be \( \frac{a}{r}, a, ar \), where \( a \) is the middle term and \( r \) is the common ratio.

2. **Use Vieta's Formulas**:
According to Vieta's formulas, for a cubic equation \( x^3 + bx^2 + cx + d = 0 \):
- The sum of the roots \( \left( \frac{a}{r} + a + ar \right) = -\frac{b}{a} \)
- The product of the roots \( \left( \frac{a}{r} \cdot a \cdot ar \right) = -\frac{d}{a} \)

For our equation:
- \( b = -13 \)
- \( d = -27 \)
- \( a = 1 \)

3. **Calculate the Product of the Roots**:
\[ \frac{a}{r} \cdot a \cdot ar = -\frac{-27}{1} = 27 \]
This simplifies to:
\[ a^3 = 27 \implies a = 3 \]

4. **Calculate the Sum of the Roots**:
\[ \frac{a}{r} + a + ar = -\frac{-13}{1} = 13 \]
Substituting \( a = 3 \):
\[ \frac{3}{r} + 3 + 3r = 13 \]
Rearranging gives:
\[ \frac{3}{r} + 3r = 13 - 3 = 10 \]
Multiplying through by \( r \) to eliminate the fraction:
\[ 3 + 3r^2 = 10r \]
Rearranging gives:
\[ 3r^2 - 10r + 3 = 0 \]

5. **Solve the Quadratic Equation**:
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
\[ = \frac{10 \pm \sqrt{100 - 36}}{6} \]
\[ = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} \]
This gives two possible values for \( r \):
\[ r = \frac{18}{6} = 3 \quad \text{or} \quad r = \frac{2}{6} = \frac{1}{3} \]

6. **Calculate \( k \)**:
Now we use the relationship for the sum of the products of the roots taken two at a time:
\[ \frac{a}{r} \cdot a + a \cdot ar + ar \cdot \frac{a}{r} = k \]
This simplifies to:
\[ a^2 \left( \frac{1}{r} + r + 1 \right) = k \]
Substituting \( a = 3 \):
\[ 3^2 \left( \frac{1}{r} + r + 1 \right) = k \]
For \( r = 3 \):
\[ k = 9 \left( \frac{1}{3} + 3 + 1 \right) = 9 \left( \frac{1}{3} + \frac{9}{3} + \frac{3}{3} \right) = 9 \left( \frac{13}{3} \right) = 39 \]
For \( r = \frac{1}{3} \):
\[ k = 9 \left( 3 + \frac{1}{\frac{1}{3}} + 1 \right) = 9 \left( 3 + 3 + 1 \right) = 9 \cdot 7 = 63 \]
However, since we are looking for \( k \) that matches the options given, we find \( k = 39 \).

### Final Answer:
Thus, the value of \( k \) is \( \boxed{39} \).