The sum of four consecutive numbers in an $AP$ is $32$ and the ratio of the product of the first and the last terms to the product of the two middle terms is $7: 15$. Find the numbers.

(2, 6, 10, 14) Let the four consecutive numbers in $AP$ be $(a-3d), (a-d), (a+d), (a+3d)$.
Given that their sum is $32$:
$(a-3d) + (a-d) + (a+d) + (a+3d) = 32$
$4a = 32$
$a = 8$
Given the ratio of the product of the first and last terms to the product of the middle terms is $7:15$:
$\frac{(a-3d)(a+3d)}{(a-d)(a+d)} = \frac{7}{15}$
$\frac{a^2 - 9d^2}{a^2 - d^2} = \frac{7}{15}$
Substitute $a = 8$:
$\frac{64 - 9d^2}{64 - d^2} = \frac{7}{15}$
$15(64 - 9d^2) = 7(64 - d^2)$
$960 - 135d^2 = 448 - 7d^2$
$512 = 128d^2$
$d^2 = 4$
$d = \pm 2$
If $a = 8$ and $d = 2$,the numbers are $(8-6), (8-2), (8+2), (8+6)$,which are $2, 6, 10, 14$.
If $a = 8$ and $d = -2$,the numbers are $(8+6), (8+2), (8-2), (8-6)$,which are $14, 10, 6, 2$.