Найдите первый член, знаменатель и количество членов конечной геометрической прогрессии (yn), если y4-y2=-24, y3+y2=6, а сумма всех членов Sn=-182.

\[\left\{ \begin{matrix} y_{4} - y_{2} = - 24 \\ y_{3} + y_{2} = 6\ \ \ \ \ \ \\ S_{n} = - 182\ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} y_{1}q^{3} - yq = - 24\ \ \ \ \\ y_{1}q^{2} + yq = 6\ \ \ \ \ \ \ \ \ \ \\ \frac{y_{1}\left( q^{4} - 1 \right)}{q - 1} = - 182 \\ \end{matrix}\text{\ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} y_{1}q\left( q^{2} - 1 \right) = - 24 \\ y_{1}q(q + 1) = 6\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[q - 1 = - 4\]

\[q = - 3.\]

\[y_{1} = \frac{- 24}{q^{3} - q} = \frac{- 24}{- 27 + 3} = 1\]

\[\frac{1 \cdot (( - 3)^{n} - 1)}{- 4} = - 182\]

\[( - 3)^{n} - 1 = 728\ \]

\[( - 3)^{n} = 729\]

\[( - 3)^{n} = ( - 3)^{6},\]

\[n = 6.\]

\[Ответ:\ \ y_{1} = 1;\ \ \ q = - 3;\ \ \ n = 6.\]