Question

Sol. [Key Concept: First A.M. $p$ and first H.M. $q$ between two numbers; relate $p$ and $q$.]

$$\begin{matrix} p = a + n\left( \frac{b - a}{n + 1} \right) = \frac{nb + a}{n + 1} \\ \frac{1}{q} = \frac{1}{a} + n\left( \frac{\frac{1}{b} - \frac{1}{a}}{n + 1} \right) = \frac{na + b}{(n + 1)ab} \\ q = \frac{(n + 1)ab}{na + b} \\ \ q\left( na + \frac{p(n + 1) - a}{n} \right) = (n + 1)a\left( \frac{(n + 1)p - a}{n} \right) \\ \therefore\ q\left( \left( n^{2} - 1 \right)a + p(n + 1) \right) = (n + 1)a((n + 1)p - a) \\ \Rightarrow \ (q - p)^{2}n^{2} + (q + p)^{2} - 2n\left( q^{2} - p^{2} \right) - 4pq \geq 0 \\ \Rightarrow \ (q - p)^{2}n^{2} + (q - p)^{2} - 2n\left( q^{2} - p^{2} \right) \geq 0 \\ \Rightarrow \ (q - p)\left( (q - p)n^{2} + (q - p) - 2n(q + p) \right) \geq 0 \\ \Rightarrow \ (q - p)\left( q(n - 1)^{2} - p(n + 1)^{2} \right) \geq 0 \\ \ a^{2} + a(q(n - 1) - (n + 1)p) + pq = 0 \\ \ q \in ( - \infty,p\rbrack \cup \left\lbrack \left( \frac{n + 1}{n - 1} \right)p,\infty \right) \end{matrix}$$