Let $$A_1, A_2, A_3, \ldots$$ be an increasing geometric progression of positive real numbers. If $$A_1 A_3 A_5 A_7 = \frac{1}{1296}$$ and $$A_2 + A_4 = \frac{7}{36}$$, then the value of $$A_6 + A_8 + A_{10}$$ is equal to

Let the increasing GP have first term $$a$$ and common ratio $$r$$ (with $$r > 1$$ since the GP is increasing and consists of positive reals). The general term is $$A_n = ar^{n-1}$$. The product $$A_1 A_3 A_5 A_7 = a \cdot ar^2 \cdot ar^4 \cdot ar^6 = a^4 r^{12} = \frac{1}{1296}$$ and the sum $$A_2 + A_4 = ar + ar^3 = ar(1 + r^2) = \frac{7}{36}$$.

From $$a^4 r^{12} = \frac{1}{1296}$$ we have $$(ar^3)^4 = \frac{1}{1296}$$, so $$ar^3 = \left(\frac{1}{1296}\right)^{1/4} = \frac{1}{6}$$ (taking the positive fourth root since all terms are positive). Hence $$a = \frac{1}{6r^3}$$.

Substituting into the second equation $$ar(1 + r^2) = \frac{7}{36}$$ gives $$\frac{1}{6r^3} \cdot r \cdot (1 + r^2) = \frac{7}{36}$$, which simplifies to $$\frac{1 + r^2}{6r^2} = \frac{7}{36}$$. Multiplying both sides by 36 yields $$36(1 + r^2) = 42r^2$$, and thus $$6r^2 = 36$$ leading to $$r^2 = 6$$.

To find $$A_6 + A_8 + A_{10}$$, note that $$A_6 + A_8 + A_{10} = ar^5 + ar^7 + ar^9 = ar^5(1 + r^2 + r^4)$$. Since $$r^2 = 6$$, we have $$1 + r^2 + r^4 = 1 + 6 + 36 = 43$$, and $$ar^5 = ar^3 \cdot r^2 = \frac{1}{6} \times 6 = 1$$. Therefore $$A_6 + A_8 + A_{10} = 1 \times 43 = 43$$.

Hence the correct answer is Option A: $$43$$.