1. Set up the condition for the A.P.

Let the A.P. have:

• first term $a=100$
• last term $l=199$
• common difference $d\in \mathbb{Z}^+$
• number of terms $n$

For an A.P., $l=a+(n-1)d$ So, $199=100+(n-1)d$ $99=(n-1)d$

Thus, $d$ must be a positive divisor of $99$, and $n=\frac{99}{d}+1.$

2. Apply the condition on number of terms

We need $3\le n\le 33.$ Using $n=\frac{99}{d}+1$:

Lower bound:

$\frac{99}{d}+1\ge 3$ $\frac{99}{d}\ge 2$ $d\le 49.5$ So $d\le 49$.

Upper bound:

$\frac{99}{d}+1\le 33$ $\frac{99}{d}\le 32$ $d\ge \frac{99}{32}$ So $d\ge 4$ since $d$ is an integer.

Therefore, $d$ must be a divisor of $99$ satisfying $4\le d\le 49.$

3. List all positive divisors of $99$

Since $99=3^2\cdot 11,$ its positive divisors are $1,3,9,11,33,99.$

Among these, those in the range $4\le d\le 49$ are: $9,11,33.$

4. Check the corresponding number of terms

• For $d=9$: $n=\frac{99}{9}+1=11+1=12$
• For $d=11$: $n=\frac{99}{11}+1=9+1=10$
• For $d=33$: $n=\frac{99}{33}+1=3+1=4$

All satisfy $3\le n\le 33$.

5. Sum the common differences

$9+11+33=53$

Therefore, the required sum is $\boxed{53}.$