ГДЗ по алгебре 9 класс Шыныбеков раздел 3 - 3.80

**1)** Дано: $b_5 = -6$, $b_7 = -54$. Найдем знаменатель прогрессии $q$, используя формулу $n$-го члена $b_n = b_k \cdot q^{n-k}$: $$ q^2 = \frac{b_7}{b_5} = \frac{-54}{-6} = 9 $$ Отсюда $q = 3$ или $q = -3$. Так как $b_5 = b_1 \cdot q^4$, а $q^4 = (\pm 3)^4 = 81$, первый член равен: $$ b_1 = \frac{b_5}{q^4} = \frac{-6}{81} = -\frac{2}{27} $$ Найдем сумму первых 6 членов по формуле $S_n = \frac{b_1(q^n - 1)}{q - 1}$: * Если $q = 3$: $$ S_6 = \frac{-\frac{2}{27}(3^6 - 1)}{3 - 1} = \frac{-\frac{2}{27}(729 - 1)}{2} = -\frac{1}{27} \cdot 728 = -\frac{728}{27} = -26\frac{26}{27} $$ * Если $q = -3$: $$ S_6 = \frac{-\frac{2}{27}((-3)^6 - 1)}{-3 - 1} = \frac{-\frac{2}{27}(729 - 1)}{-4} = \frac{728}{54} = \frac{364}{27} = 13\frac{13}{27} $$ **Ответ:** $-\frac{728}{27}$ при $q = 3$; $\frac{364}{27}$ при $q = -3$. --- **2)** Дано: $b_6 = 25$, $b_8 = 9$. Найдем $q^2$: $$ q^2 = \frac{b_8}{b_6} = \frac{9}{25} \implies q = \frac{3}{5} \quad \text{или} \quad q = -\frac{3}{5} $$ * Случай 1: $q = \frac{3}{5}$. Найдем $b_1$: $$ b_1 = \frac{b_6}{q^5} = \frac{25}{\left(\frac{3}{5}\right)^5} = \frac{25}{\frac{243}{3125}} = \frac{78125}{243} $$ Вычислим $S_6$: $$ S_6 = \frac{b_1(1 - q^6)}{1 - q} = \frac{\frac{78125}{243}\left(1 - \frac{729}{15625}\right)}{1 - \frac{3}{5}} = \frac{\frac{78125}{243} \cdot \frac{14896}{15625}}{\frac{2}{5}} = \frac{5 \cdot 14896}{243} \cdot \frac{5}{2} = \frac{186200}{243} = 766\frac{62}{243} $$ * Случай 2: $q = -\frac{3}{5}$. Найдем $b_1$: $$ b_1 = \frac{b_6}{q^5} = \frac{25}{\left(-\frac{3}{5}\right)^5} = -\frac{78125}{243} $$ Вычислим $S_6$: $$ S_6 = \frac{b_1(1 - q^6)}{1 - q} = \frac{-\frac{78125}{243}\left(1 - \frac{729}{15625}\right)}{1 - \left(-\frac{3}{5}\right)} = \frac{-\frac{78125}{243} \cdot \frac{14896}{15625}}{\frac{8}{5}} = -\frac{5 \cdot 14896}{243} \cdot \frac{5}{8} = -\frac{46550}{243} = -191\frac{137}{243} $$ **Ответ:** $\frac{186200}{243}$ при $q = \frac{3}{5}$; $-\frac{46550}{243}$ при $q = -\frac{3}{5}$. --- **3)** Дано: $b_1 = 125$, $b_3 = 5$. Найдем $q^2$: $$ q^2 = \frac{b_3}{b_1} = \frac{5}{125} = \frac{1}{25} \implies q = \frac{1}{5} \quad \text{или} \quad q = -\frac{1}{5} $$ * Если $q = \frac{1}{5}$: $$ S_6 = \frac{b_1(1 - q^6)}{1 - q} = \frac{125\left(1 - \left(\frac{1}{5}\right)^6\right)}{1 - \frac{1}{5}} = \frac{125\left(1 - \frac{1}{15625}\right)}{\frac{4}{5}} = \frac{125 \cdot \frac{15624}{15625}}{\frac{4}{5}} = \frac{15624}{125} \cdot \frac{5}{4} = \frac{15624}{100} = 156,24 $$ * Если $q = -\frac{1}{5}$: $$ S_6 = \frac{b_1(1 - q^6)}{1 - q} = \frac{125\left(1 - \left(-\frac{1}{5}\right)^6\right)}{1 - \left(-\frac{1}{5}\right)} = \frac{125\left(1 - \frac{1}{15625}\right)}{\frac{6}{5}} = \frac{125 \cdot \frac{15624}{15625}}{\frac{6}{5}} = \frac{15624}{125} \cdot \frac{5}{6} = \frac{2604}{25} = 104,16 $$ **Ответ:** $156,24$ при $q = \frac{1}{5}$; $104,16$ при $q = -\frac{1}{5}$. --- **4)** Дано: $b_4 = -1$, $b_6 = -100$. Найдем $q^2$: $$ q^2 = \frac{b_6}{b_4} = \frac{-100}{-1} = 100 \implies q = 10 \quad \text{или} \quad q = -10 $$ * Если $q = 10$: $$ b_1 = \frac{b_4}{q^3} = \frac{-1}{10^3} = -\frac{1}{1000} = -0,001 $$ $$ S_6 = \frac{b_1(q^6 - 1)}{q - 1} = \frac{-0,001(10^6 - 1)}{10 - 1} = \frac{-0,001 \cdot 999999}{9} = -0,001 \cdot 111111 = -111,111 $$ * Если $q = -10$: $$ b_1 = \frac{b_4}{q^3} = \frac{-1}{(-10)^3} = \frac{-1}{-1000} = \frac{1}{1000} = 0,001 $$ $$ S_6 = \frac{b_1(q^6 - 1)}{q - 1} = \frac{0,001((-10)^6 - 1)}{-10 - 1} = \frac{0,001(1000000 - 1)}{-11} = \frac{999,999}{-11} = -90,909 $$ **Ответ:** $-111,111$ при $q = 10$; $-90,909$ при $q = -10$.