Arithmetic Progression: General Term

Let, $$d_a$$ and $$d_b$$ be the common differences of the two APs.

We have, 

$$ b_{43}-b_{31} = (b_1+42d_b) - (b_1+30d_b) = 12d_b $$

$$ \Rightarrow 12d_b =-385 - (-277) = -108 \Rightarrow d_b = -9 $$

And, $$ d_a = d_b + 13 = -9 + 13 = 4 $$

Now, $$ a_{78} = a_1 + 77d_a = 327 $$

$$\Rightarrow a_1 + 77 \cdot 4 = 327$$

$$ \Rightarrow a_1 = 327 - 308 = 19 $$

We are given that $$a, b, \frac{1}{18}$$ are in a geometric progression and that $$\frac{1}{a}, 10, \frac{1}{b}$$ are in an arithmetic progression, and we need to find $$16a + 12b$$.

Because three terms in geometric progression satisfy that the square of the middle term equals the product of the outer terms, we have

$$b^2 = a \times \frac{1}{18} = \frac{a}{18}$$

which implies $$a = 18b^2$$.

Similarly, three terms in arithmetic progression satisfy that twice the middle term equals the sum of the outer terms, so

$$2 \times 10 = \frac{1}{a} + \frac{1}{b}$$

and therefore $$\frac{1}{a} + \frac{1}{b} = 20$$.

Substituting $$a = 18b^2$$ into this equation leads to $$\frac{1}{18b^2} + \frac{1}{b} = 20$$. Multiplying both sides by $$18b^2$$ gives

$$1 + 18b = 360b^2$$

or equivalently

$$360b^2 - 18b - 1 = 0$$

Applying the quadratic formula yields

$$b = \frac{18 \pm \sqrt{324 + 1440}}{720} = \frac{18 \pm \sqrt{1764}}{720} = \frac{18 \pm 42}{720}$$

Since $$a,b > 0$$, we select the positive root $$b = \frac{60}{720} = \frac{1}{12}$$.

Substituting $$b = \frac{1}{12}$$ into $$a = 18b^2$$ gives $$a = 18 \times \frac{1}{144} = \frac{1}{8}$$.

Finally, calculating the required expression, we find

$$16a + 12b = 16 \times \frac{1}{8} + 12 \times \frac{1}{12} = 2 + 1 = 3$$

Hence, the answer is $$3$$.

The sequence $$A$$ is an arithmetic progression with first term $$1$$ and common difference $$5$$.
The sequence $$B$$ is an arithmetic progression with first term $$9$$ and common difference $$7$$.

Both sets contain exactly 2025 terms.
We must find the last term of the sequence that ends first to determine the upper boundary for their intersection.

The last term of set $$A$$ is:
$$L_A = 1 + (2025 - 1) \times 5 = 1 + 2024 \times 5 = 10121$$

The last term of set $$B$$ is:
$$L_B = 9 + (2025 - 1) \times 7 = 9 + 2024 \times 7 = 14177$$

The intersection of sets $$A$$ and $$B$$ contains common terms that form a new arithmetic progression.
The first common term is 16.
The common difference for this intersection sequence is the least common multiple of the original common differences, which is $$\text{LCM}(5, 7) = 35$$.

Let the number of common terms be $$m$$. The general term for the common sequence is $$16 + (m - 1) \times 35$$.

This term must be less than or equal to the smaller of the two final terms, which is 10121.

$$16 + (m - 1) \times 35 \le 10121$$
$$35(m - 1) \le 10105$$
$$m - 1 \le \frac{10105}{35}$$
$$m - 1 \le 288.71$$

Since $$m$$ must be an integer, the maximum value for $$m - 1$$ is 288.
$$m = 289$$
This means the number of common terms $$n(A \cap B)$$ is 289.

To find the union of the two sets, we apply the principle of inclusion and exclusion:
$$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$
$$n(A \cup B) = 2025 + 2025 - 289$$
$$n(A \cup B) = 4050 - 289 = 3761$$

Hence the correct option is C.