Find four numbers in $A.P.$ such that their sum is $36$ and the product of means ($2^{nd}$ and $3^{rd}$ term) exceeds the product of extremes ($1^{st}$ and $4^{th}$ term) by $32.$

(A) Let the four numbers in $A.P.$ be $(a-3d, a-d, a+d, a+3d).$
According to the first condition,the sum of the numbers is $36$:
$(a-3d) + (a-d) + (a+d) + (a+3d) = 36$
$4a = 36$
$a = 9$
According to the second condition,the product of the means exceeds the product of the extremes by $32$:
$(a-d)(a+d) = (a-3d)(a+3d) + 32$
$a^2 - d^2 = a^2 - 9d^2 + 32$
$8d^2 = 32$
$d^2 = 4$
$d = \pm 2$
Case $1$: If $a = 9$ and $d = 2$,the numbers are $(9-6, 9-2, 9+2, 9+6) = (3, 7, 11, 15).$
Case $2$: If $a = 9$ and $d = -2$,the numbers are $(9+6, 9+2, 9-2, 9-6) = (15, 11, 7, 3).$
Thus,the required four numbers are $3, 7, 11, 15$ or $15, 11, 7, 3.$