1. Write the two progressions explicitly as APs

   • First progression: $4,9,14,19,\ldots$

     This is an AP with: $a_1=4,\quad d_1=5$ Its $n$-th term is $T_n=4+(n-1)5=5n-1$

   • Second progression: $3,6,9,12,\ldots$

     This is an AP with: $a_2=3,\quad d_2=3$ Its $m$-th term is $S_m=3+(m-1)3=3m$

2. Find the last term included in each progression

   • Up to the $25^{\text{th}}$ term of the first AP: $T_{25}=4+24\cdot 5=124$

   • Up to the $37^{\text{th}}$ term of the second AP: $S_{37}=3\cdot 37=111$

   So common terms must be numbers that appear in both APs and are therefore $\le 111$.

3. Condition for a common term

   A common term must satisfy: $x\equiv 4\pmod{5}$ and also $x\equiv 0\pmod{3}$

   Let $x=3k$. Then: $3k\equiv 4\pmod{5}$ Since $3^{-1}\equiv 2\pmod{5}$, $k\equiv 8\equiv 3\pmod{5}$ Hence: $k=5t+3$ Therefore, $x=3(5t+3)=15t+9$

   So the common terms form the sequence: $9,24,39,54,69,84,99,114,\ldots$

4. Restrict to the given ranges

   We need common terms that are present in both lists.

   Since the second list goes only up to $111$, we take common terms $\le 111$: $9,24,39,54,69,84,99$

   The next one is $114>111$, so it is not included.

   Thus, the number of common terms is: $7$

5. Check with options

   Option $D$ is $7$.

Final Answer: $\boxed{7}$