In the following $APs,$ find the missing terms in the boxes $5, \square, \square, 9 \frac{1}{2}$

(N/A) Given the $A.P.$ as $5, \square, \square, 9 \frac{1}{2}$.
Here,the first term $a = 5$.
The fourth term $a_4 = 9 \frac{1}{2} = \frac{19}{2}$.
We know the formula for the $n^{th}$ term of an $A.P.$ is $a_n = a + (n - 1)d$.
For $n = 4$,$a_4 = a + 3d$.
Substituting the values: $\frac{19}{2} = 5 + 3d$.
Subtracting $5$ from both sides: $\frac{19}{2} - 5 = 3d \implies \frac{19 - 10}{2} = 3d \implies \frac{9}{2} = 3d$.
Dividing by $3$,we get the common difference $d = \frac{3}{2}$.
Now,find the missing terms:
The second term $a_2 = a + d = 5 + \frac{3}{2} = \frac{10 + 3}{2} = \frac{13}{2}$.
The third term $a_3 = a + 2d = 5 + 2(\frac{3}{2}) = 5 + 3 = 8$.
Thus,the missing terms are $\frac{13}{2}$ and $8$.