Решебник по алгебре 9 класс Мерзляк Задание 889

\[\boxed{\mathbf{889\ (889).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[q = 3;\ \ c_{n} = 162;\ \ S_{n} = 242\]

\[c_{n} = 162:\ \]

\[c_{n} = c_{1} \cdot q^{n - 1}\text{\ \ }\]

\[c_{1} \cdot q^{n - 1} = 162\]

\[\frac{c_{1} \cdot 3^{n}}{3} = 162\ \ \]

\[c_{1} \cdot 3^{n} = 162 \cdot 3\ \]

\[c_{1} \cdot 3^{n} = 486\]

\[S_{n} = 242:\]

\[S_{n} = \frac{c_{1} \cdot \left( q^{n} - 1 \right)}{q - 1} = 242\ \ \]

\[\frac{c_{1}\left( 3^{n} - 1 \right)}{3 - 1} = 242\]

\[c_{1}\left( 3^{n} - 1 \right) = 484\ \]

\[\left\{ \begin{matrix} c_{1} \cdot 3^{n} = 486\ \ \ \ \ \ \ \ \\ c_{1}\left( 3^{n} - 1 \right) = 484 \\ \end{matrix} \right.\ \ \ |\ \ (:)\ \]

\[\ \frac{c_{1} \cdot 3^{n}}{c_{1}\left( 3^{n} - 1 \right)} = \frac{486}{484}\text{\ \ }\]

\[\frac{3^{n}}{3^{n - 1}} = \frac{243}{242},\]

\[242 \cdot 3^{n} = 243 \cdot \left( 3^{n} - 1 \right)\]

\[242 \cdot 3^{n} = 243 \cdot 3^{n} - 243\]

\[242 \cdot 3^{n} - 243 \cdot 3^{n} =\]

\[= - 243\ \ | \cdot ( - 1)\]

\[3^{n} \cdot (243 - 242) = 243\]

\[3^{n} \cdot 1 = 243\]

\[3^{n} = 3^{5} \Longrightarrow \ \ n = 5\]

\[Ответ:n = 5.\]