In an increasing geometric series, the sum of the second and the sixth term is $$\frac{25}{2}$$ and the product of the third and fifth term is 25. Then, the sum of $$4^{th}, 6^{th}$$ and $$8^{th}$$ terms is equal to:

Let the geometric series have first term $$a$$ and common ratio $$r > 1$$ (since the series is increasing). The terms are $$a, ar, ar^2, ar^3, \ldots$$

We are given that the sum of the second and sixth terms is $$\frac{25}{2}$$: $$ar + ar^5 = \frac{25}{2}$$, and the product of the third and fifth terms is 25: $$ar^2 \cdot ar^4 = a^2 r^6 = 25$$.

From the product condition, $$a^2 r^6 = 25$$, so $$ar^3 = 5$$ (taking the positive root since terms are positive in an increasing GP). This gives $$a = \frac{5}{r^3}$$.

Substituting into the sum condition: $$\frac{5}{r^3} \cdot r + \frac{5}{r^3} \cdot r^5 = \frac{25}{2}$$, which simplifies to $$\frac{5}{r^2} + 5r^2 = \frac{25}{2}$$, so $$r^2 + \frac{1}{r^2} = \frac{5}{2}$$.

Let $$u = r^2$$. Then $$u + \frac{1}{u} = \frac{5}{2}$$, giving $$2u^2 - 5u + 2 = 0$$, so $$(2u - 1)(u - 2) = 0$$. Since $$r > 1$$, we need $$r^2 = u = 2$$.

The sum of the 4th, 6th, and 8th terms is $$ar^3 + ar^5 + ar^7 = ar^3(1 + r^2 + r^4) = 5(1 + 2 + 4) = 35$$.