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 :

Given: GP $$a, ar, ar^2 \dots$$ (increasing, positive).

1. Product of 3rd and 5th: $$ar^2 \cdot ar^4 = 49 \implies a^2r^6 = 49 \implies ar^3 = 7$$.
2. Sum of 2nd and 6th: $$ar + ar^5 = \frac{70}{3}$$.
3. Find terms: $$ar^3 = 7$$.
4. Sum: $$7 + 21 + 63 = \mathbf{91}$$. (Option B)

Substitute $$a = \frac{7}{r^3}$$: $$\frac{7}{r^3}r + \frac{7}{r^3}r^5 = \frac{70}{3} \implies \frac{7}{r^2} + 7r^2 = \frac{70}{3}$$.

Let $$x = r^2$$: $$\frac{1}{x} + x = \frac{10}{3} \implies 3x^2 - 10x + 3 = 0$$.

$$(3x-1)(x-3) = 0 \implies x = 3$$ (since it's increasing, $$r^2 > 1$$). So $$r^2 = 3$$.

$$4^{th} \text{ term} = ar^3 = 7$$.

$$6^{th} \text{ term} = ar^5 = (ar^3)r^2 = 7 \cdot 3 = 21$$.

$$8^{th} \text{ term} = ar^7 = (ar^5)r^2 = 21 \cdot 3 = 63$$.