\[\boxed{\text{618\ (618).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[x_{1} = 17,\ \ x_{2} = 14,\ \ d = x_{2} - x_{1} = 14 - 17 = - 3:\]
\[S_{n} = \frac{2x_{1} + d(n - 1)}{2} \cdot n = \frac{34 - 3 \cdot (n - 1)}{2} \cdot n > 0 \Longrightarrow\]
\[\Longrightarrow 34n - 3n^{2} + 3n > 0 \Longrightarrow 3n^{2} - 37n < 0 \Longrightarrow\]
\[n(3n - 37) < 0\]
\[n_{1} = 0,\ \ n_{2} = 12\frac{1}{3};\]
\[так\ как\ n \in N \Longrightarrow n = 12.\]
\[Ответ:\ \ 12\ членов.\ \]
\[\boxed{\text{618.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ b_{n} = 0,2 \cdot 5^{n},\]
\[\ \ q = \frac{b_{n}}{b_{n - 1}} = \frac{0,2 \cdot 5^{n}}{0,2 \cdot 5^{n - 1}} = 5,\]
\[\text{\ \ }b_{1} = 0,2 \cdot 5 = 1,\]
\[\Longrightarrow b_{n} - геометрическая\]
\[\ прогрессия;\]
\[S_{n} = b_{1} \cdot \frac{q^{n} - 1}{q - 1} = \frac{5^{n} - 1}{5 - 1} =\]
\[= \frac{5^{n} - 1}{4};\]
\[\textbf{б)}\ b_{n} = 3 \cdot 2^{n - 1},\]
\[\ \ q = \frac{b_{n}}{b_{n - 1}} = \frac{3 \cdot 2^{n - 1}}{3 \cdot 2^{n - 2}} = 2,\]
\[\text{\ \ }b_{1} = 3\]
\[\Longrightarrow b_{n} - геометрическая\ \]
\[прогрессия;\]
\[S_{n} = b_{1} \cdot \frac{q^{n} - 1}{q - 1} = \frac{3 \cdot \left( 2^{n} - 1 \right)}{2 - 1} =\]
\[= 3 \cdot (2^{n} - 1);\]
\[\textbf{в)}\ b_{n} = 3^{1 + n},\]
\[\ \ q = \frac{b_{n}}{b_{n - 1}} = \frac{3^{1 + n}}{3^{n}} = 3,\ \ \]
\[b_{1} = 9\]
\[\Longrightarrow b_{n} - геометрическая\ \]
\[прогрессия;\]
\[S_{n} = b_{1} \cdot \frac{q^{n} - 1}{q - 1} = 9 \cdot \frac{3^{n} - 1}{3 - 1} =\]
\[= \frac{9}{2} \cdot (3^{n} - 1).\]
\[\textbf{г)}\ b_{n} = 2^{n + 2};\ \ \ q = \frac{b_{n}}{b_{n - 1}\ } =\]
\[= \frac{2^{n + 2}}{2^{n + 1}} = 2;\ \ \ \ b_{1} = 8\]
\[\Longrightarrow b_{n} - геометрическая\ \]
\[прогрессия;\]
\[S_{n} = b_{1} \cdot \frac{q^{n} - 1}{q - 1} = 8 \cdot \frac{2^{n} - 1}{2 - 1} =\]
\[= 8 \cdot \left( 2^{n} - 1 \right).\]