Find five numbers in $A.P.$ such that their sum is $30$ and the sum of their squares is $220$.

(N/A) Let the required five numbers in $A.P.$ be $a-2d, a-d, a, a+d, a+2d$.
By the first condition,
$(a-2d) + (a-d) + a + (a+d) + (a+2d) = 30$
$5a = 30$
$a = 6$
By the second condition,
$(a-2d)^2 + (a-d)^2 + a^2 + (a+d)^2 + (a+2d)^2 = 220$
$(a^2 - 4ad + 4d^2) + (a^2 - 2ad + d^2) + a^2 + (a^2 + 2ad + d^2) + (a^2 + 4ad + 4d^2) = 220$
$5a^2 + 10d^2 = 220$
$a^2 + 2d^2 = 44$
Substituting $a = 6$:
$6^2 + 2d^2 = 44$
$36 + 2d^2 = 44$
$2d^2 = 8$
$d^2 = 4$
$d = \pm 2$
If $a = 6$ and $d = 2$,the numbers are $6-4, 6-2, 6, 6+2, 6+4$,which are $2, 4, 6, 8, 10$.
If $a = 6$ and $d = -2$,the numbers are $6+4, 6+2, 6, 6-2, 6-4$,which are $10, 8, 6, 4, 2$.
Thus,the five numbers are $2, 4, 6, 8, 10$ or $10, 8, 6, 4, 2$.