Let the three numbers in A.P. be $a-d,\:a,\:a+d$
Given that the sum of the numbers $=24$
$\Rightarrow\:a-d+a+a+d=24$
$\Rightarrow\:3a=24$ or $a=8$
Also given that the product of the numbers $=440$
$\Rightarrow\:(a-d).a(a+d)=440$
$\Rightarrow\:a(a^2-d^2)=440$
Substituting the value of $a$ we get
$8(64-d^2)=440$
$\Rightarrow\:64-d^2=\large\frac{440}{8}$$=55$
$\Rightarrow\:d^2=64-55=9$ or $d=\pm3$
Substituting the value of $a$ and $d$ we get the three numbers are
$5,8\:11$