1. Write the two arithmetic progressions explicitly

   • Set $A$ consists of the first $2025$ terms of the AP $1,6,11,16,\ldots$ with first term $a_1=1$ and common difference $d_1=5$.

     So $A=\{1+5k\mid k=0,1,2,\ldots,2024\}.$

   • Set $B$ consists of the first $2025$ terms of the AP $9,16,23,30,\ldots$ with first term $b_1=9$ and common difference $d_2=7$.

     So $B=\{9+7m\mid m=0,1,2,\ldots,2024\}.$

2. Use inclusion-exclusion

   We need $n(A\cup B)=n(A)+n(B)-n(A\cap B).$

   Since each set has $2025$ elements, $n(A\cup B)=2025+2025-n(A\cap B)=4050-n(A\cap B).$

   So the main task is to find $n(A\cap B)$.

3. Find common terms of the two APs

   A number common to both sets must satisfy $x\equiv 1\pmod{5}$ and $x\equiv 9\pmod{7}.$

   Since $9\equiv 2\pmod{7}$, this is $x\equiv 1\pmod{5},\qquad x\equiv 2\pmod{7}.$

   Let us solve this congruence.

   Write $x=1+5k.$ Then $1+5k\equiv 2\pmod{7}$ $5k\equiv 1\pmod{7}.$

   Since the inverse of $5$ modulo $7$ is $3$ (because $5\cdot 3=15\equiv 1\pmod{7}$), we get $k\equiv 3\pmod{7}.$

   So $k=3+7t$ and hence $x=1+5(3+7t)=16+35t.$

   Therefore the common terms form an AP: $A\cap B=\{16,51,86,121,\ldots\}$ with common difference $35$.

4. Count how many common terms lie in both first 2025 terms

   Since both sets are truncated to their first $2025$ terms, a common term must not exceed the last term of either AP.

   • Last term of $A$: $1+(2025-1)\cdot 5=1+2024\cdot 5=1+10120=10121.$

   • Last term of $B$: $9+(2025-1)\cdot 7=9+2024\cdot 7=9+14168=14177.$

   So a common term must be at most $10121$ (the smaller upper bound).

   Thus we count terms of $16+35t\le 10121.$

   Solve: $35t\le 10121-16=10105$ $t\le \frac{10105}{35}=288.714\ldots$

   Hence $t=0,1,2,\ldots,288.$

   Number of such integers is $288-0+1=289.$

   Therefore, $n(A\cap B)=289.$

5. Apply inclusion-exclusion

   $n(A\cup B)=4050-289=3761.$

6. Check options

   • A: $3814$ ❌
   • B: $4003$ ❌
   • C: $4027$ ❌
   • D: $3761$ ✅

Therefore, the correct answer is $\boxed{3761}.$