Given: The sequence is in H.P. with \(a_m = n\) and \(a_n = m\) \((m \ne n)\).

In an H.P., reciprocals of terms form an A.P. So, \(\tfrac{1}{a_m} = \tfrac{1}{n}\) and \(\tfrac{1}{a_n} = \tfrac{1}{m}\).

This A.P. leads to common difference \(d = \tfrac{1}{mn}\) and first term \(A = \tfrac{1}{mn}\).

Thus, the reciprocal of the \((m+n)\)th term is:

\(b_{m+n} = \tfrac{m+n}{mn}\)

Hence, the \((m+n)\)th term of H.P. is:

\(a_{m+n} = \dfrac{mn}{m+n}\)

Given: Mean of squares of first n natural numbers is 11.

Use the identity for sum of squares: $$\sum_{k=1}^{n} k^2=\frac{n(n+1)(2n+1)}{6}$$

Mean \(=\) (sum) \( \div n \Rightarrow \) $$\frac{1}{n}\sum_{k=1}^{n}k^2=\frac{(n+1)(2n+1)}{6}=11$$

Solve for \(n\): $$(n+1)(2n+1)=66 $$ $$\;\Rightarrow\; 2n^2+3n+1=66 \;\Rightarrow\; $$ $$n=\frac{-3\pm\sqrt{9+520}}{4}=\frac{-3\pm 23}{4}$$

Only positive integer solution: $$n=\frac{20}{4}=5$$

Answer: \(n=5\).

Equation:

\(6x^2 - xy - 2y^2 = 0\)

Compare with: \( ax^2 + 2hxy + by^2 = 0 \)

\(a = 6\)

\(2h = -1 \;\;\Rightarrow\;\; h = -\tfrac{1}{2}\)

\(b = -2\)

Slopes satisfy:

\(bm^2 + 2hm + a = 0\)

\(-2m^2 - m + 6 = 0\)

\(\Rightarrow 2m^2 + m - 6 = 0\)

Solving quadratic:

\(m = \dfrac{-1 \pm \sqrt{49}}{4}\)

\(m = \dfrac{-1 \pm 7}{4}\)

Therefore:

\(m_1 = \tfrac{3}{2}, \quad m_2 = -2\)