Question

Sol. [Key Concept: From $4m + 1$ tickets choose 3 in A.P.: count common differences $d = 1..m$.]

Let number $s$ are $x_{1},x_{1} + 2d,x_{1} + 4d$

$$\begin{matrix} & x_{1} + 4d \leq 4m + 1 \\ & \ \Rightarrow x_{1} \leq 4(m - d) + 1,d = 1,2,3,\ldots,m \end{matrix}$$

∴ Number of favorable ways $= (4m - 3) + (4m - 7) + \ldots.. + 1$

$$\begin{matrix} & \ = \frac{m}{2}(4m - 2) = m(2m - 1) \\ & \text{~}\text{Number of total ways}\text{~} = \ ^{4m + 1}C_{3} = \frac{(4m + 1)4m(4m - 1)}{6} \\ & \ P = \frac{m(2m - 1)6}{(4m + 1)4m(4m - 1)} = \frac{3(2m - 1)}{2\left( 16m^{2} - 1 \right)} \end{matrix}$$