Let a, b and c be in G.P. with common ratio r, where $$a \neq 0$$ and $$0 \lt r \leq \frac{1}{2}$$. If 3a, 7b and 15c are the first three terms of an A.P., then the 4$$^{th}$$ term of this A.P. is:

We have that $$a,\;b,\;c$$ are in geometric progression with common ratio $$r$$, so by definition of a G.P.

$$b = ar \quad \text{and} \quad c = ar^{2}.$$

The extra information $$0 \lt r \le \dfrac12$$ will be used later to select the admissible value of $$r$$.

Now the numbers $$3a,\;7b,\;15c$$ form an arithmetic progression. For any A.P. the common difference is the same between every pair of consecutive terms, i.e.

$$\bigl(7b-3a\bigr) = \bigl(15c-7b\bigr).$$

Substituting $$b = ar$$ and $$c = ar^{2}$$ gives

$$7(ar) - 3a \;=\; 15(ar^{2}) - 7(ar).$$

Because $$a \neq 0$$ we can divide the entire equation by $$a$$ without changing the equality:

$$7r - 3 \;=\; 15r^{2} - 7r.$$

Bringing every term to the right produces the quadratic equation

$$15r^{2} - 7r - 7r + 3 = 0 \quad\Longrightarrow\quad 15r^{2} - 14r + 3 = 0.$$

We solve this quadratic using the formula $$r = \dfrac{-B \pm \sqrt{B^{2} - 4AC}}{2A}$$ for $$Ax^{2}+Bx+C=0$$. Here $$A = 15,\;B = -14,\;C = 3$$, so

$$r = \dfrac{14 \pm \sqrt{(-14)^{2} - 4 \cdot 15 \cdot 3}}{2 \cdot 15} = \dfrac{14 \pm \sqrt{196 - 180}}{30} = \dfrac{14 \pm \sqrt{16}}{30} = \dfrac{14 \pm 4}{30}.$$

This yields two possible roots:

$$r_{1} = \dfrac{14 + 4}{30} = \dfrac{18}{30} = \dfrac35 \quad\text{and}\quad r_{2} = \dfrac{14 - 4}{30} = \dfrac{10}{30} = \dfrac13.$$

Because we are told $$0 \lt r \le \dfrac12$$, the value $$r = \dfrac35$$ (which equals 0.6) is inadmissible. Hence we must take

$$r = \dfrac13.$$

Substituting this value back, the G.P. terms become

$$b = a\left(\dfrac13\right) = \dfrac{a}{3}, \qquad c = a\left(\dfrac13\right)^{2} = \dfrac{a}{9}.$$

The corresponding A.P. terms are therefore

$$T_{1} = 3a, \qquad T_{2} = 7b = 7\left(\dfrac{a}{3}\right) = \dfrac{7a}{3}, \qquad T_{3} = 15c = 15\left(\dfrac{a}{9}\right) = \dfrac{5a}{3}.$$

The common difference $$d$$ of the A.P. is

$$d = T_{2} - T_{1} = \dfrac{7a}{3} - 3a = \dfrac{7a - 9a}{3} = -\dfrac{2a}{3}.$$

Alternatively, $$d = T_{3} - T_{2} = \dfrac{5a}{3} - \dfrac{7a}{3} = -\dfrac{2a}{3},$$ confirming the same value.

The fourth term of an A.P. is given by $$T_{4} = T_{3} + d$$. Substituting the values just obtained:

$$T_{4} = \dfrac{5a}{3} + \left(-\dfrac{2a}{3}\right) = \dfrac{5a - 2a}{3} = \dfrac{3a}{3} = a.$$

Hence, the correct answer is Option A.