The sum of three numbers in $A.P.$ is $48$ and the sum of their squares is $800$. Find those numbers.

(A) Let the three numbers in $A.P.$ be $(a-d)$,$a$,and $(a+d)$.
According to the first condition:
$(a-d) + a + (a+d) = 48$
$3a = 48$
$a = 16$
According to the second condition:
$(a-d)^2 + a^2 + (a+d)^2 = 800$
$(a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) = 800$
$3a^2 + 2d^2 = 800$
Substituting $a = 16$:
$3(16)^2 + 2d^2 = 800$
$3(256) + 2d^2 = 800$
$768 + 2d^2 = 800$
$2d^2 = 32$
$d^2 = 16$
$d = \pm 4$
Case $1$: If $a = 16$ and $d = 4$,the numbers are $(16-4), 16, (16+4)$,which are $12, 16, 20$.
Case $2$: If $a = 16$ and $d = -4$,the numbers are $(16-(-4)), 16, (16+(-4))$,which are $20, 16, 12$.
Thus,the required numbers are $12, 16, 20$ or $20, 16, 12$.