Se $a_1 = 2$ e $a_2 = 1$ , $\large{\frac{a_2}{a_1} = q = \frac{1}{2}}$ (razão da progressão) Logo, o termo enésimo é $a_n = 2\cdot2^{1-n} = 2^{2-n}$ , assim $a_8 = \large{\frac{1}{64}}$ . $S = \log_2{2} + \log_2{1} + \log_2{\frac{1}{2}} + \cdots + \log_2{\frac{1}{64}} $ $=$ $\log_2{2\cdot 1\cdot \frac{1}{2} \cdots \frac{1}{64}}$ $S = \log_2{\frac{1}{4} \cdot \frac{1}{8} \cdots \frac{1}{64}}$ $=$ $\log_2{2^{-2} \cdot 2^{-3} \cdot 2^{-4} \cdot 2^{-5} \cdot 2^{-6}}$ $=$ $\log_2{2^{-20}}$ $S = -20$ $\implies$ $\color{red}{b = - \Large{\frac{S}{5}}}$ $\color{red}{=}$ $\color{red}{4}$ $f(x) = |x+8|+|2x-4|$ $\implies$ $\boxed{f(1) = |9|+|-2| = 11}$