Question

6А.27,6А.29 и 6А.30 пожалуйста

Answer 1

6А.27
$b_n=b_1q^{n-1} \\\ a) \ b_7=4\cdot ( \frac{1}{2} )^{7-1}=4\cdot ( \frac{1}{2} )^6=4\cdot \frac{1}{64}=\frac{1}{16} \\\ b) \ b_5= \frac{1}{5} \cdot (-2)^{5-1}= \frac{1}{5} \cdot (-2)^{5-1}= \frac{1}{5} \cdot (-2)^4= \frac{16}{5}=3.2$

6A.29
$S= \frac{b_1}{1-q} =\cfrac{b_1}{1- \frac{b_2}{b_1} } \\\ a) \ S= \cfrac{25}{1- \frac{5}{25} } = \cfrac{25}{1- \frac{1}{5} } =\cfrac{25}{ \frac{4}{5} } = \frac{125}{4} =31 \frac{1}{4} \\\ b) \ S= \cfrac{10}{1- \frac{1}{10} } = \cfrac{10}{ \frac{9}{10} } = \frac{100}{9} =11 \frac{1}{9}$

6A.30
$a) \ a_n= \frac{(-1)^n}{9n-10} \\\ a_1= \frac{(-1)^1}{9\cdot1-10} = \frac{-1}{9-10} =\frac{-1}{-1} =1 \\\ a_2= \frac{(-1)^2}{9\cdot2-10} =\frac{1}{18-10} =\frac{1}{8} \\\ a_3= \frac{(-1)^3}{9\cdot3-10} =\frac{-1}{27-10} =-\frac{1}{17} \\\ a_4= \frac{(-1)^4}{9\cdot4-10} =\frac{1}{36-10} =\frac{1}{26} \\\ a_5= \frac{(-1)^5}{9\cdot5-10} = \frac{-1}{45-10}=- \frac{1}{35} \\\ a_6= \frac{(-1)^6}{9\cdot6-10} = \frac{1}{54-10} = \frac{1}{44}$
$b) \ a_n= \frac{(-1)^{n+1}}{5n+2} \\\ a_1= \frac{(-1)^{1+1}}{5\cdot1+2} = \frac{(-1)^2}{5+2} = \frac{1}{7} \\\ a_2= \frac{(-1)^{2+1}}{5\cdot2+2} = \frac{(-1)^3}{10+2} =- \frac{1}{12} \\\ a_3= \frac{(-1)^{3+1}}{5\cdot3+2} = \frac{(-1)^4}{15+2} = \frac{1}{17} \\\ a_4= \frac{(-1)^{4+1}}{5\cdot4+2} = \frac{(-1)^5}{20+2} =- \frac{1}{22} \\\ a_5= \frac{(-1)^{5+1}}{5\cdot5+2} = \frac{(-1)^6}{25+2} = \frac{1}{27} \\\ a_6= \frac{(-1)^{6+1}}{5\cdot6+2} = \frac{(-1)^7}{30+2} =- \frac{1}{32}$