If the roots of the equation $x^{3}-lx^{2}+mx-n=0$ are in H.P., then the mean root is
- $\displaystyle \frac{3n}{m}$
- $\displaystyle \frac{2n}{m}$
- $\displaystyle \frac{n}{m}$
- $-\displaystyle \frac{n}{m}$
If roots a, b, c are in H.P., then 1/a, 1/b, 1/c are in A.P. For a cubic equation x^3 - lx^2 + mx - n = 0, the product of roots is n. If the roots are in H.P., the middle root is 3n/m.
If the roots of x^3 - lx^2 + mx - n = 0 are in harmonic progression, their reciprocals are in arithmetic progression, meaning the roots can be written as 1/(a-d), 1/a, and 1/(a+d). The product of the roots is n, so (1/(a-d)) * (1/a) * (1/(a+d)) = 1 / (a*(a^2 - d^2)) = n, which implies a*(a^2 - d^2) = 1/n. The sum of the pairwise products is m, so calculating 1/(a*(a-d)) + 1/(a*(a+d)) + 1/(a^2 - d^2) and simplifying yields m = 3 / (a*(a^2 - d^2)). Substituting 1/n into this relationship gives m = 3n, and solving for the mean root 1/a gives the result 3n/m.