Let $$A = \{1, 6, 11, 16, \ldots\}$$ and $$B = \{9, 16, 23, 30, \ldots\}$$ be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $$n(A \cup B)$$ is

Set $$A$$ is an arithmetic progression with first term $$1$$ and common difference $$5$$.
Its $$k^{\text{th}}$$ term is $$a_k = 1 + (k-1)\cdot5 = 5k - 4\; (1 \le k \le 2025)$$.

Set $$B$$ is an arithmetic progression with first term $$9$$ and common difference $$7$$.
Its $$m^{\text{th}}$$ term is $$b_m = 9 + (m-1)\cdot7 = 7m + 2\; (1 \le m \le 2025)$$.

We want $$n(A \cup B)$$. The formula for two finite sets is
$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$.

Because each progression contains the first $$2025$$ terms,
$$n(A) = 2025,\; n(B) = 2025$$.

So we must determine $$n(A \cap B)$$, the count of common elements.

An element is common when
$$5k - 4 = 7m + 2 \quad -(1)$$
for some integers $$k,m$$ in the range $$1$$ to $$2025$$.

Rewriting $$(1)$$:
$$5k - 7m = 6 \quad -(2)$$

Equation $$(2)$$ is linear Diophantine. Solve it first without bounds.

Work modulo $$7$$:
$$5k \equiv 6 \pmod 7$$.
The inverse of $$5$$ modulo $$7$$ is $$3$$ (since $$5\cdot3 \equiv 1 \pmod 7$$).
Thus $$k \equiv 6\cdot3 = 18 \equiv 4 \pmod 7$$.

Write $$k = 4 + 7t$$ for some integer $$t$$. Substitute into $$(2)$$:

$$5(4 + 7t) - 7m = 6$$
$$20 + 35t - 7m = 6$$
$$7m = 14 + 35t$$
$$m = 2 + 5t$$.

Hence the complete solution set is
$$k = 4 + 7t,\;\; m = 2 + 5t \quad -(3)$$
with $$t$$ an integer.

Apply the bounds $$1 \le k,m \le 2025$$.

From $$(3)$$,
$$k = 4 + 7t \ge 1 \;\Rightarrow\; t \ge 0$$ (since $$t$$ is integer).
Also $$4 + 7t \le 2025 \;\Rightarrow\; 7t \le 2021 \;\Rightarrow\; t \le 288$$.

For these $$t$$ values, $$m = 2 + 5t$$ runs from $$2$$ (when $$t = 0$$) to
$$2 + 5\cdot288 = 1442$$ (when $$t = 288$$), which is still within $$1 \le m \le 2025$$, so no further restriction occurs.

Therefore $$t$$ can take every integer from $$0$$ to $$288$$ inclusive, giving

$$n(A \cap B) = 288 - 0 + 1 = 289$$.

Finally,

$$n(A \cup B) = 2025 + 2025 - 289 = 3761$$.

Thus the number of distinct elements in $$A \cup B$$ is $$3761$$, which is Option C.