If the roots of 10x³ - nx² - 54x - 27 = 0 are in harmonic progression, then n equals ___________.

 $10{\mathrm{x}}^3-{\mathrm{nx}}^2-54\mathrm{x}-27=0$ has roots in H.P.

put x = 1/t

$27{\mathrm{t}}^3+54{\mathrm{t}}^2+\mathrm{nt}-10=0$

This equation has roots in A.P. Let the roots be a - d, a, and a + d

$\therefore 3\mathrm{a}=-\frac{54}{27}\text{ or }\mathrm{a}=-\frac{2}{3}$

Also $(\mathrm{a}-\mathrm{d})\mathrm{a}(\mathrm{a}+\mathrm{d})=\frac{10}{27}$

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\frac{2}{3}\left(\frac{4}{9}-{\mathrm{d}}^2\right)=-\frac{10}{27}\text{ or }\left(\frac{4}{9}-{\mathrm{d}}^2\right)=-\frac{5}{9}\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{d}}^2=1\text{ or }\mathrm{d}=\pm 1\end{array}$

So, roots are $\frac{1}{3},-\frac{2}{3},-\frac{5}{3}$

$\therefore \frac{\mathrm{n}}{27}=\frac{10}{9}-\frac{5}{9}-\frac{2}{9}\text{ or }\frac{\mathrm{n}}{27}=\frac{3}{9}$

or n = 9