Страница 216 номер 724, ГДЗ по алгебре за 9 класс к учебнику Мерзляка
Найдите формулу \(n\)-го члена арифметической прогрессии: 1) \(-5,\ -7,\ -9,\ -11,\ \ldots\); 2) \(2,\ 2\frac{1}{6},\ 2\frac{1}{3},\ 2\frac{1}{2},\ \ldots\); 3) \(a^2,\ 2a^2,\ 3a^2,\ 4a^2,\ \ldots\); 4) \(a + 3,\ a + 1,\ a - 1,\ a - 3,\ \ldots\) .
1) \(- 5, - 7, - 9, - 11,\ldots;\)
\(a_{1} = - 5\)
\(d = - 7 - ( - 5) = - 2\)
\(a_{n} = a_{1} + d(n - 1) = - 5 - 2(n - 1) = - 5 - 2n + 2 = - 3 - 2n\)
2) \(2,\ 2\frac{1}{6},2\frac{1}{3},\ 2\frac{1}{2},\ \ldots;\)
\(a_{1} = 2\)
\(d = 2\dfrac{1}{6} - 2 = \dfrac{1}{6}\)
\(a_{n} = a_{1} + d(n - 1) = 2 + \dfrac{1}{6}(n - 1) = 2 + \dfrac{1}{6}n - \dfrac{1}{6} = 1\dfrac{5}{6} + \dfrac{1}{6}n\)
3) \(a^{2},2a^{2},3a^{2},4a^{2},\ldots;\)
\(a_{1} = a^{2}\)
\(d = 2a^{2} - a^{2} = a^{2}\)
\(a_{n} = a_{1} + d(n - 1) = a^{2} + a^{2}(n - 1) = a^{2} + a^{2} \cdot n - a^{2} = na^{2}\)
4) \(a + 3,a + 1,a - 1,a - 3,\ldots\)
\(a_{1} = a + 3\)
\(d = a + 1 - (a + 3) = a + 1 - a - 3 = - 2\)
\(a_{n} = a_{1} + d(n - 1) = a + 3 - 2(n - 1) =\)
\(= a + 3 - 2n + 2 = a + 5 - 2n\)
Ответ: 1) \(a_{n} = - 3 - 2n\); 2) \(a_{n} = 1\frac{5}{6} + \frac{1}{6}n\); 3) \(a_{n} = na^{2}\); 4) \(a_{n} = a + 5 - 2n\).
