The sum of the $4^{th}$ and $8^{th}$ terms of an $AP$ is $24$ and the sum of the $6^{th}$ and $10^{th}$ terms is $44$. Find the first three terms of the $AP$.

(D) We know that the $n^{th}$ term of an $AP$ is given by $a_n = a + (n - 1)d$.
For the $4^{th}$ and $8^{th}$ terms:
$a_4 = a + 3d$
$a_8 = a + 7d$
Given $a_4 + a_8 = 24$,we have $(a + 3d) + (a + 7d) = 24$,which simplifies to $2a + 10d = 24$ or $a + 5d = 12$ $...(1)$
For the $6^{th}$ and $10^{th}$ terms:
$a_6 = a + 5d$
$a_{10} = a + 9d$
Given $a_6 + a_{10} = 44$,we have $(a + 5d) + (a + 9d) = 44$,which simplifies to $2a + 14d = 44$ or $a + 7d = 22$ $...(2)$
Subtracting equation $(1)$ from $(2)$:
$(a + 7d) - (a + 5d) = 22 - 12$
$2d = 10$
$d = 5$
Substituting $d = 5$ into equation $(1)$:
$a + 5(5) = 12$
$a + 25 = 12$
$a = -13$
The first three terms are:
$a_1 = a = -13$
$a_2 = a + d = -13 + 5 = -8$
$a_3 = a + 2d = -13 + 10 = -3$
Thus,the first three terms of the $AP$ are $-13, -8, -3$.