Let $$a_1, a_2, a_3, \ldots a_n, \ldots$$, be in A.P. If $$a_3 + a_7 + a_{11} + a_{15} = 72$$, then the sum of its first 17 terms is equal to:

Let us denote the first term of the arithmetic progression by $$a_1$$ and its common difference by $$d$$.

For an A.P. the formula for the $$k^{\text{th}}$$ term is $$a_k = a_1 + (k-1)d$$. We now write each term that appears in the given condition.

We have

$$\begin{aligned} a_3 &= a_1 + (3-1)d = a_1 + 2d$$, $$\\[4pt] a_7 &= a_1 + (7-1)d = a_1 + 6d$$, $$\\[4pt] a_{11} &= a_1 + (11-1)d = a_1 + 10d$$, $$\\[4pt] a_{15} &= a_1 + (15-1)d = a_1 + 14d. \end{aligned}$$

Adding these four expressions term-by-term, we get

$$\begin{aligned} a_3 + a_7 + a_{11} + a_{15} &= (a_1 + 2d) + (a_1 + 6d) + (a_1 + 10d) + (a_1 + 14d)\\[4pt] &= 4a_1 + (2 + 6 + 10 + 14)d\\[4pt] &= 4a_1 + 32d. \end{aligned}$$

The question states that this sum equals $$72$$, hence

$$4a_1 + 32d = 72.$$

Dividing every term by $$4$$, we simplify to obtain

$$a_1 + 8d = 18.$$

Our next goal is to find the sum of the first $$17$$ terms. The formula for the sum of the first $$n$$ terms of an A.P. is stated as $$S_n = \dfrac{n}{2}\,[\,2a_1 + (n-1)d\,]$$. Here $$n = 17$$, so we write

$$S_{17} = \frac{17}{2}\,[\,2a_1 + 16d\,].$$

We already have a relation between $$a_1$$ and $$d$$ from $$a_1 + 8d = 18$$. Let us solve this relation for $$a_1$$:

$$a_1 = 18 - 8d.$$

Substituting this value of $$a_1$$ into the bracketed expression $$2a_1 + 16d$$, we obtain

$$\begin{aligned} 2a_1 + 16d &= 2\,(18 - 8d) + 16d\\[4pt] &= 36 - 16d + 16d\\[4pt] &= 36. \end{aligned}$$

Now we substitute this result back into the sum formula:

$$\begin{aligned} S_{17} &= \frac{17}{2}\,(36)\\[4pt] &= 17 \times 18\\[4pt] &= 306. \end{aligned}$$

Hence, the correct answer is Option A.