Let the coefficients of three consecutive terms in the binomial expansion of $$(1 + 2x)^n$$ be in the ratio $$2 : 5 : 8$$. Then the coefficient of the term, which is in the middle of these three terms, is

Three consecutive terms of $$(1 + 2x)^n$$ have coefficients in the ratio $$2 : 5 : 8$$. Since the general term is $$T_{r+1} = \binom{n}{r}(2x)^r = \binom{n}{r} \cdot 2^r \cdot x^r$$, the coefficient of $$x^r$$ is $$\binom{n}{r} \cdot 2^r$$.

For three consecutive terms with indices $$r, r+1, r+2$$, the ratio of the coefficient of the $$(r+1)$$th term to the $$r$$th term gives $$\frac{\binom{n}{r+1} \cdot 2^{r+1}}{\binom{n}{r} \cdot 2^r} = \frac{5}{2} \implies \frac{n-r}{r+1} \cdot 2 = \frac{5}{2} \implies \frac{n-r}{r+1} = \frac{5}{4},$$ which leads to $$4(n-r) = 5(r+1) \implies 4n = 9r + 5 \quad \cdots (i).$$

Next, the ratio of the coefficient of the $$(r+2)$$th term to the $$(r+1)$$th term yields $$\frac{\binom{n}{r+2} \cdot 2^{r+2}}{\binom{n}{r+1} \cdot 2^{r+1}} = \frac{8}{5} \implies \frac{n-r-1}{r+2} \cdot 2 = \frac{8}{5} \implies \frac{n-r-1}{r+2} = \frac{4}{5},$$ which gives $$5(n-r-1) = 4(r+2) \implies 5n = 9r + 13 \quad \cdots (ii).$$

Subtracting $$(i)$$ from $$(ii)$$ gives $$n = 8$$, and substituting back into $$(i)$$ yields $$32 = 9r + 5 \implies r = 3$$.

Therefore the middle term, corresponding to index $$r+1 = 4$$, has coefficient $$\binom{8}{4} \cdot 2^4 = 70 \times 16 = 1120.$$ The correct answer is 1120.