Even a set can be sincere, why not you?

Let five consecutive integers are like $a-2,a-1,a,a+1,a+2$ .Its sum is $5a$

If $5a$ should ends with $5$ then $a$ should be odd integer.

Here $a$ should starts with $3$ and ends with $97$ because minimum of $a-2=1\,\,i.e.\,\, a=3$ and maximum of $a+2=99\,\,i.e.\,\, a=97$ .

So $3 \leq a \leq 97$ .

Total number of set of five consecutive integers less than 100 is $95$ because it starts with $(1,2,3,4,5)$ and ends with $95.96.97,98,99)$ .

No of value of $a=48$ and total number of sets of five consecutive integers is $95$ .

Then probability is $\dfrac{48}{95}$

Pick the maximum possible sum of the possible 5-tuples which is $95 + 96 + 97 + 98 + 99 = 485$ .
Let $x - 2, x - 1, x , x + 1, x + 2$ be a group of sincere positive integers.
$\implies (x - 2) + (x - 1) + x + (x + 1) + (x + 2) \leq 485\\ \implies x \leq 97$ .

First try $(x - 2) + (x - 1) + x + (x + 1) + (x + 2) = 5\\ x = 1\\ \\ (x - 2) + (x - 1) + x + (x+ 1) + (x + 2) = 15\\ x = 3\\ \\ (x - 2) + (x - 1) + x + (x + 1) + (x + 2) = 25\\ x = 5\\ \\ (x - 2) + (x - 1) + x + (x + 1) + (x + 2) = 35\\ \\ x = 7$ .

So we observe the sets are sincere $\forall a0x \in$ AP $1,3,5, \cdots$ .

But when, $x = 1$ , $x - 2 = -1$ and $x - 1 = 0$ which is not possible since $x$ is a positive inetegr.

So, the number of values of $x$ for which set is sincere is the number representing which term 97 is of the AP $3,5,7,...$ which is $48$ .

And the total number of 5-tubles consisting of consecutive positive integers less than 100 is $95$ .

$\therefore \dfrac{a}{b} = \dfrac{48}{95}\\ \implies a + b = 48 + 95 = \boxed{143}$ .