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(D) Let the $n$-th term of the $GP$ be $a_n = a r^{n-1}$.Given that each term is the arithmetic mean of the next two terms:$a_n = \frac{a_{n+1} + a_{n

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$-2^{15}$

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(D) Let the $n$-th term of the $GP$ be $a_n = a r^{n-1}$.Given that each term is the arithmetic mean of the next two terms:$a_n = \frac{a_{n+1} + a_{n+2}}{2}$$2 a r^{n-1} = a r^n + a r^{n+1}$Dividing by $a r^{n-1}$ (since $a 
eq 0$ and $r 
eq 0$):$2 = r + r^2$$r^2 + r - 2 = 0$$(r + 2)(r - 1) = 0$Since $a_2 
eq a_1$,we have $r 
eq 1$,so $r = -2$.We need to find $S_{20} - S_{18} = a_{19} + a_{20}$.$a_{19} + a_{20} = a r^{18} + a r^{19} = a r^{18}(1 + r)$.Substituting $a = \frac{1}{8} = 2^{-3}$ and $r = -2$:$S_{20} - S_{18} = 2^{-3} (-2)^{18} (1 - 2)$$S_{20} - S_{18} = 2^{-3} (2^{18}) (-1)$$S_{20} - S_{18} = -2^{15}$.

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

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