In an increasing geometric progression of positive terms, the sum of the second and sixth terms is $$\frac{70}{3}$$ and the product of the third and fifth terms is 49. Then the sum of the $$4^{th}$$, $$6^{th}$$ and $$8^{th}$$ terms is equal to :

Let the increasing GP of positive terms have first term $$a$$ and common ratio $$r > 1$$.

The terms are: $$a, ar, ar^2, ar^3, ar^4, ar^5, ar^6, ar^7, \ldots$$

Sum of 2nd and 6th terms = $$\frac{70}{3}$$

$$ar + ar^5 = \frac{70}{3}$$

Product of 3rd and 5th terms = 49

$$ar^2 \cdot ar^4 = a^2r^6 = 49$$

$$\Rightarrow (ar^3)^2 = 49 \Rightarrow ar^3 = 7$$ (since positive terms)$$

From $$ar^3 = 7$$, we get $$a = \frac{7}{r^3}$$.

Substituting into Condition 1:

$$\frac{7}{r^3} \cdot r + \frac{7}{r^3} \cdot r^5 = \frac{70}{3}$$

$$\frac{7}{r^2} + 7r^2 = \frac{70}{3}$$

$$7\left(\frac{1}{r^2} + r^2\right) = \frac{70}{3}$$

$$\frac{1}{r^2} + r^2 = \frac{10}{3}$$

Let $$t = r^2 + \frac{1}{r^2}$$, so $$t = \frac{10}{3}$$.

Let $$u = r^2$$: $$u + \frac{1}{u} = \frac{10}{3}$$

$$3u^2 - 10u + 3 = 0$$

$$(3u - 1)(u - 3) = 0$$

$$u = 3 \text{ or } u = \frac{1}{3}$$

Since $$r > 1$$, we need $$r^2 = 3$$, so $$r = \sqrt{3}$$.

Then $$a = \frac{7}{(\sqrt{3})^3} = \frac{7}{3\sqrt{3}}$$.

Now find the sum of 4th, 6th, and 8th terms:

$$ar^3 + ar^5 + ar^7 = ar^3(1 + r^2 + r^4) = 7(1 + 3 + 9) = 7 \times 13 = 91$$

The correct answer is Option 2: 91.