Question

Sol. [Key Concept: $p$th, $q$th, $r$th terms of an H.P. ⇒ reciprocals in A.P.; the cyclic combination telescopes to $0$.]

$$\frac{1}{x} = a + (p - 1)d\ \frac{1}{y} = a + (q - 1)d\ \frac{1}{z} = a + (r - 1)d$$

$$\begin{matrix} & \frac{1}{x} - \frac{1}{y} = (p - q)d \\ \Rightarrow & xy(p - q) = \frac{y - x}{d};yz(q - r) = \frac{z - y}{d};zx(r - p) = \frac{x - z}{d} \\ \Rightarrow & (p - q)xy + yz(q - r) + zx(r - p) = \frac{1}{d}(y - x + z - y + x - z) = 0 \end{matrix}$$