Insert 4 A.M.s between 4 and 19.

Let $A_{1}, A_{2}, A_{3}, A_{4}$ be the four A.M.s between 4 and 19. Then, 4, $A_{1}, A_{2}, A_{3}, A_{4}$ and 19 are in A.P. whose common difference is as follows:

$d=\frac{19-4}{4+1}$

= 3

$A_{1}=4+d=4+3=7$

$A_{2}=4+2 d=4+6=10$

$A_{3}=4+3 d=4+9=13$

$A_{4}=4+4 d=4+12=16$

Hence, the required A.M.s are 7, 10, 13, 16.