If the roots of the equation $10x^3 - cx^2 - 54x - 27 = 0$ are in harmonic progression, then the value of $c$ must be equal to
Step-by-Step Solution
Key Concept: If roots are in harmonic progression, replacing $x$ by $\frac{1}{x}$ transforms the equation to one whose roots are in arithmetic progression.
The roots of $10x^3 - cx^2 - 54x - 27 = 0$ are in HP. If the roots are in HP, let them be $\frac{1}{a-d}, \frac{1}{a}, \frac{1}{a+d}$. Then replacing $x$ by $\frac{1}{x}$ gives $27x^3 - 54x^2 - cx - 10 = 0$. The roots of this equation are in AP: $a-d, a, a+d$. From the sum of roots: $3a = \frac{54}{27} = 2$, so $a = \frac{2}{3}$. From the product of roots: $(a-d) \cdot a \cdot (a+d) = \frac{10}{27}$, giving $a(a^2-d^2) = \frac{10}{27}$. Solving: $\frac{2}{3}(\frac{4}{9}-d^2) = \frac{10}{27}$, which yields $d = \frac{2}{3}$. Therefore, $c = 9$.
Correct Answer: 9