Let $$a, b, c > 1$$, $$a^3, b^3$$ and $$c^3$$ be in A.P. and $$\log_a b$$, $$\log_c a$$ and $$\log_b c$$ be in G.P. If the sum of first 20 terms of an A.P., whose first term is $$\frac{a+4b+c}{3}$$ and the common difference is $$\frac{a-8b+c}{10}$$ is $$-444$$, then $$abc$$ is equal to

We are given $$a, b, c > 1$$ with $$a^3, b^3, c^3$$ in A.P. and $$\log_a b, \log_c a, \log_b c$$ in G.P.

From the G.P. condition, $$(\log_c a)^2 = \log_a b \cdot \log_b c$$. By the chain rule of logarithms, $$\log_a b \cdot \log_b c = \log_a c$$, so $$(\log_c a)^2 = \log_a c$$. Let $$t = \log_a c$$. Then $$\log_c a = \dfrac{1}{t}$$, and hence $$\dfrac{1}{t^2} = t$$, leading to $$t^3 = 1$$. Since $$a, c > 1$$ implies $$t > 0$$, we have $$t = 1$$, which gives $$\log_a c = 1$$ and therefore $$c = a$$.

With $$c = a$$, the arithmetic progression condition $$2b^3 = a^3 + c^3 = 2a^3 \implies b^3 = a^3 \implies b = a$$. Thus all three variables are equal, and we write $$a = b = c = k$$.

The first term of the new arithmetic progression is $$\dfrac{a + 4b + c}{3} = \dfrac{k + 4k + k}{3} = 2k$$, and its common difference is $$\dfrac{a - 8b + c}{10} = \dfrac{k - 8k + k}{10} = \dfrac{-6k}{10} = \dfrac{-3k}{5}$$.

Using the formula for the sum of the first 20 terms, we have
$$S_{20} = \dfrac{20}{2}\left[2(2k) + 19\left(\dfrac{-3k}{5}\right)\right] = 10\left[4k - \dfrac{57k}{5}\right]$$
$$= 10 \cdot \dfrac{20k - 57k}{5} = 10 \cdot \dfrac{-37k}{5} = -74k$$. Since $$S_{20} = -444$$, it follows that $$-74k = -444$$ and therefore $$k = 6$$.

Finally, $$abc = k^3 = 6^3 = 216$$, so the correct answer is Option B: 216.