(18, 8) The given sequence is $\square, 13, \square, 3$.
For this $AP$,the second term is $a_2 = 13$ and the fourth term is $a_4 = 3$.
We know that the $n^{th}$ term of an $AP$ is given by $a_n = a + (n - 1)d$,where $a$ is the first term and $d$ is the common difference.
For $a_2 = 13$:
$a + (2 - 1)d = 13 \implies a + d = 13$ $...(i)$
For $a_4 = 3$:
$a + (4 - 1)d = 3 \implies a + 3d = 3$ $...(ii)$
Subtracting equation $(i)$ from equation $(ii)$:
$(a + 3d) - (a + d) = 3 - 13$
$2d = -10$
$d = -5$
Substituting $d = -5$ into equation $(i)$:
$a + (-5) = 13$
$a = 13 + 5 = 18$
Now,the first term is $a = 18$ and the third term is $a_3 = a + 2d$:
$a_3 = 18 + 2(-5) = 18 - 10 = 8$.
Therefore,the missing terms are $18$ and $8$ respectively.