In the following $APs,$ find the missing terms in the boxes: $-4, \square, \square, \square, \square, 6$

(N/A) For this $A.P.,$ the first term $a = -4$ and the sixth term $a_{6} = 6$.
We know the formula for the $n^{th}$ term of an $A.P.$ is $a_{n} = a + (n - 1)d$.
Substituting the values for the sixth term:
$a_{6} = a + (6 - 1)d$
$6 = -4 + 5d$
$10 = 5d$
$d = 2$
Now,we find the missing terms:
$a_{2} = a + d = -4 + 2 = -2$
$a_{3} = a + 2d = -4 + 2(2) = 0$
$a_{4} = a + 3d = -4 + 3(2) = 2$
$a_{5} = a + 4d = -4 + 4(2) = 4$
Therefore,the missing terms are $-2, 0, 2,$ and $4$ respectively.