Страница 230 номер 836, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Найдите первый член геометрической прогрессии \((c_n)\), если: 1) \(c_4 = \frac{1}{98}\), а знаменатель \(q = \frac{2}{7}\); 2) \(c_6 = 100\), \(c_9 = 100\,000\).
\(c_{n} = c_{1}q^{n - 1} \Rightarrow c_{1} = \dfrac{c_{n}}{q^{n - 1}}\)
1) \(c_{4} = \frac{1}{98}\), \(q = \frac{2}{7}\)
\(c_{1} = \dfrac{c_{4}}{q^{3}} = \dfrac{1}{98} \cdot \left( \dfrac{7}{2} \right)^{3} = \dfrac{1}{98} \cdot \dfrac{7^{3}}{2^{3}} = \dfrac{1}{2 \cdot 7^{2}} \cdot \dfrac{7^{3}}{2^{3}} = \dfrac{7}{2^{4}} = \dfrac{7}{16}\)
2) \(c_{6} = 100\), \(c_{9} = 100\ 000\)
\(c_{1} = \dfrac{c_{6}}{q^{5}} = \dfrac{c_{9}}{q^{8}}\)
\(\dfrac{c_{9}}{c_{6}} = q^{3} \Rightarrow q^{3} = \dfrac{100000}{100} = 1000 \Rightarrow q = 10\)
\(c_{1} = \dfrac{100}{10^{5}} = \dfrac{10^{2}}{10^{5}} = 10^{- 3}\)
Ответ: 1) \(c_{1} = \frac{7}{16}\); 2) \(c_{1} = 10^{- 3}\).
