From a month of 31 days, 3 different dates are selected at random....

The total number of ways to select $3$ different dates from a month of $31$ days is given by choosing $3$ days out of $31$:

$^{31}C_{3} = \dfrac{31 \times 30 \times 29}{3 \times 2 \times 1} = 4495$

Let the three selected dates be $x, y, z$ such that $x < y < z$. For these dates to form an increasing Arithmetic Progression (A.P.), they must satisfy the condition:

$2y = x + z$

For $y$ to be an integer, the sum $x + z$ must be an even number. This is only possible if both $x$ and $z$ are either odd or even. Once $x$ and $z$ are chosen, $y$ is uniquely determined as their exact average.

In a month of $31$ days, the number of odd dates ($1, 3, 5, \dots, 31$) is $16$, and the number of even dates ($2, 4, 6, \dots, 30$) is $15$.

The number of ways to choose $x$ and $z$ such that both are odd is:

$^{16}C_{2} = \dfrac{16 \times 15}{2} = 120$

The number of ways to choose $x$ and $z$ such that both are even is:

$^{15}C_{2} = \dfrac{15 \times 14}{2} = 105$

The total number of favorable selections that form an increasing A.P. is:

$120 + 105 = 225$

The probability that the selected dates are in an increasing A.P. is:

$P = \dfrac{225}{4495}$

Dividing the numerator and the denominator by $5$, we get:

$P = \dfrac{45}{899}$

Since $45 = 3^2 \times 5$ and $899 = 29 \times 31$, they share no common factors, meaning $\gcd(45, 899) = 1$.

Thus, $a = 45$ and $b = 899$.

Finally, $a + b = 45 + 899 = 944$.

Answer: $944$