If each term of a geometric progression $$a_1, a_2, a_3, \ldots$$ with $$a_1 = \frac{1}{8}$$ and $$a_2 \neq a_1$$, is the arithmetic mean of the next two terms and $$S_n = a_1 + a_2 + \ldots + a_n$$, then $$S_{20} - S_{18}$$ is equal to

We are given a GP $$a_1, a_2, a_3, \ldots$$ with $$a_1 = \frac{1}{8}$$ and $$a_2 \neq a_1$$, where each term is the arithmetic mean of the next two terms.

Let the common ratio be $$r$$, so that $$a_n = a_1 \cdot r^{n-1}$$. The condition that each term is the arithmetic mean of the next two terms translates to $$a_n = \frac{a_{n+1} + a_{n+2}}{2}$$. Substituting $$a_n = a_1 r^{n-1}$$ yields $$a_1 r^{n-1} = \frac{a_1 r^n + a_1 r^{n+1}}{2}$$. Dividing both sides by $$a_1 r^{n-1}$$ (which is non-zero) gives $$1 = \frac{r + r^2}{2}$$, so $$2 = r + r^2$$ and hence $$r^2 + r - 2 = 0$$. Factoring, $$(r + 2)(r - 1) = 0$$, and since $$a_2 \neq a_1$$ implies $$r \neq 1$$, we conclude $$r = -2$$.

Next, to find $$S_{20} - S_{18}$$, observe that $$S_{20} - S_{18} = a_{19} + a_{20} = a_1 r^{18} + a_1 r^{19} = a_1 r^{18}(1 + r) = \frac{1}{8} \cdot (-2)^{18} \cdot (1 + (-2)) = \frac{1}{8} \cdot 2^{18} \cdot (-1) = -\frac{2^{18}}{2^3} = -2^{15}$$.

Answer: Option 4 — $$-2^{15}$$