If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Let the three numbers in A.P. be $a-d, a$, and $a+d$.

According to the given information,

$(a-d)+(a)+(a+d)=24 \ldots(1)$

$\Rightarrow 3 a=24$

$\therefore a=8$

$(a-d) a(a+d)=440$

$\Rightarrow(8-d)(8)(8+d)=440$

$\Rightarrow(8-d)(8+d)=55$

$\Rightarrow 64-d^{2}=55$

$\Rightarrow d^{2}=64-55=9$

$\Rightarrow d=\pm 3$

Therefore, when d = 3, the numbers are 5, 8, and 11 and when d = –3, the numbers are 11, 8, and 5.

Thus, the three numbers are 5, 8, and 11.