If the roots of `10 x^3-n x^2-54 x-27=0` are in harmonic oprogresion, then `n` eqauls _________.

Explanation

To solve the problem, we need to find the value of \( n \) such that the roots of the cubic equation \( 10x^3 - nx^2 - 54x - 27 = 0 \) are in harmonic progression. Here’s a step-by-step solution:

### Step 1: Understanding the Relationship Between Harmonic and Arithmetic Progression
The roots being in harmonic progression means that their reciprocals are in arithmetic progression. Therefore, if the roots are \( r_1, r_2, r_3 \), then \( \frac{1}{r_1}, \frac{1}{r_2}, \frac{1}{r_3} \) are in arithmetic progression.

### Step 2: Substitute \( x = \frac{1}{t} \)
To transform the equation, we substitute \( x = \frac{1}{t} \):
\[ 10\left(\frac{1}{t}\right)^3 - n\left(\frac{1}{t}\right)^2 - 54\left(\frac{1}{t}\right) - 27 = 0 \]
This simplifies to:
\[ \frac{10}{t^3} - \frac{n}{t^2} - \frac{54}{t} - 27 = 0 \]

### Step 3: Multiply by \( t^3 \)
To eliminate the fractions, multiply the entire equation by \( t^3 \):
\[ 10 - nt - 54t^2 - 27t^3 = 0 \]
Rearranging gives:
\[ 27t^3 + 54t^2 + nt - 10 = 0 \]

### Step 4: Roots in Arithmetic Progression
Assuming the roots are \( a - d, a, a + d \), we can express the sum of the roots:
\[ (a - d) + a + (a + d) = 3a \]
According to Vieta's formulas, the sum of the roots is:
\[ - \frac{\text{coefficient of } t^2}{\text{coefficient of } t^3} = -\frac{54}{27} = -2 \]
Thus:
\[ 3a = -2 \implies a = -\frac{2}{3} \]

### Step 5: Product of the Roots
The product of the roots is given by:
\[ (a - d) \cdot a \cdot (a + d) = a(a^2 - d^2) \]
Using Vieta's formulas, we have:
\[ - \frac{\text{constant term}}{\text{coefficient of } t^3} = -\frac{-10}{27} = \frac{10}{27} \]
Thus:
\[ a(a^2 - d^2) = \frac{10}{27} \]

### Step 6: Substitute \( a \) into the Product Equation
Substituting \( a = -\frac{2}{3} \):
\[ -\frac{2}{3} \left( \left(-\frac{2}{3}\right)^2 - d^2 \right) = \frac{10}{27} \]
Calculating \( a^2 \):
\[ \left(-\frac{2}{3}\right)^2 = \frac{4}{9} \]
Thus:
\[ -\frac{2}{3} \left( \frac{4}{9} - d^2 \right) = \frac{10}{27} \]
Multiplying both sides by -3:
\[ 2 \left( \frac{4}{9} - d^2 \right) = -\frac{30}{27} = -\frac{10}{9} \]
This simplifies to:
\[ \frac{8}{9} - 2d^2 = -\frac{10}{9} \]
Rearranging gives:
\[ 2d^2 = \frac{8}{9} + \frac{10}{9} = \frac{18}{9} = 2 \implies d^2 = 1 \]
Thus, \( d = \pm 1 \).

### Step 7: Finding the Roots
The roots are:
\[ -\frac{2}{3} - 1, -\frac{2}{3}, -\frac{2}{3} + 1 \]
Calculating gives:
\[ -\frac{5}{3}, -\frac{2}{3}, \frac{1}{3} \]

### Step 8: Sum of the Product of Roots Taken Two at a Time
Using Vieta's formulas again, the sum of the product of the roots taken two at a time is:
\[ \frac{n}{27} \]
Calculating the products:
\[ \left(-\frac{5}{3}\right)\left(-\frac{2}{3}\right) + \left(-\frac{5}{3}\right)\left(\frac{1}{3}\right) + \left(-\frac{2}{3}\right)\left(\frac{1}{3}\right) \]
Calculating each term:
1. \( \frac{10}{9} \)
2. \( -\frac{5}{9} \)
3. \( -\frac{2}{9} \)

Adding these:
\[ \frac{10}{9} - \frac{5}{9} - \frac{2}{9} = \frac{3}{9} = \frac{1}{3} \]
Thus:
\[ \frac{n}{27} = \frac{1}{3} \implies n = 9 \]

### Final Answer
The value of \( n \) is \( 9 \).

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