The total number of ways in which three distinct numbers in Arithmetic Progression can be selected from the set  $\left\{1,2,3,\mathrm{.......},24\right\}$ is equal to

$\text{ Let the numbers selected be }{x}_1,x_2,x_3$

$\text{ We must have }2x_2=x_1+x_3$

$\Rightarrow x_1+x_3=\text{ even }$

$\text{ Therefore both }{x}_1,x_3\text{ are odd (or) both are even. }$

$\text{ If }{x}_1{\text{ and x}}_3{\text{ both are even, we can select them in }}^{12}{C}_2\text{ ways. }$

$\text{ Similarly, if both }{x}_1\text{ and }{x}_3\text{ are odd, }$

${\text{ we can again select them in }}^{12}{C}_2\text{ ways. }$

$\text{ Thus, the total number of ways is }2{\times }^{12}{C}_2=132$