Question

Sol. [Key Concept: Numbers in H.P. ⇒ their reciprocals are in A.P.; use the sum and sum-of-reciprocals.]

$$(a - d) + a + (a + d) = 1 \Rightarrow a = \frac{1}{3}$$

$$\begin{matrix} & & \frac{1}{a - d} + \frac{1}{a} + \frac{1}{a + d} & = 11 \\ \Rightarrow & & 2a & = 8\left( a^{2} - d^{2} \right) \\ \Rightarrow & & & \frac{1}{3} = 4\left( \frac{1}{9} - d^{2} \right) \\ \Rightarrow & & d^{2} & = \frac{1}{36} \\ \Rightarrow & & \text{ }d & = \pm \frac{1}{6}\text{~}\text{numbers are}\text{~}6,3,2 \end{matrix}$$