Найдите первый член, знаменатель и количество членов конечной геометрической прогрессии (cn), если c6-c4=135, c6-c5=81, а сумма всех членов Sn=665.

Ответ:

\[\left\{ \begin{matrix} c_{6} - c_{4} = 135 \\ c_{6} - c_{5} = 81\ \ \\ S_{n} = 665\ \ \ \ \ \ \ \ \\ \end{matrix}\text{\ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} c_{1}q^{5} - c_{1}q^{3} = 135 \\ c_{1}q^{5} - c_{1}q^{4} = 81\ \ \\ \frac{c_{1}\left( q^{n} - 1 \right)}{q - 1} = 665\ \ \\ \end{matrix}\text{\ \ \ \ \ \ \ \ \ } \right.\ \]

\[\left\{ \begin{matrix} c_{1}q^{3}\left( q^{2} - 1 \right) = 135 \\ c_{1}q^{4}(q - 1) = 81\ \ \ \ \ \\ \frac{c_{1}(q^{n} - 1)}{q - 1} = 665\ \ \ \ \\ \end{matrix} \right.\ \]

\[\frac{q + 1}{q} = \frac{5}{3}\text{\ \ \ \ \ }\]

\[3q + 3 - 5q = 0\ \]

\[q = \frac{3}{2} = 1,5.\]

\[\frac{32 \cdot \left( \left( \frac{3}{2} \right)^{n} - 1 \right)}{\frac{1}{2}} = 665\ \ \ \]

\[\left( \frac{3}{2} \right)^{n} - 1 = \frac{665}{64}\text{\ \ \ \ }\]

\[\left( \frac{3}{2} \right)^{n} = \frac{729}{64} = \left( \frac{3}{2} \right)^{6}\]

\[n = 6.\]

\[Ответ:c_{1} = 32;\ \ \ \ q = 1,5;\ \ \ \ \ \]

\[n = 6.\]