[tex]1. \\b_n=n-\frac12n+8\\b_5=5-\frac12*5+8\\b_5=\frac{10}2-\frac52+8\\b_5=\frac{5}2+8\\b_5=2\frac12+8\\b_5=10\frac12b_5=10,5\\\\b_{n+1}=(n+1)-\frac12(n+1)+8\\b_{n+1}=n+1-\frac12n-\frac12+8\\b_{n+1}=\frac12n-\frac12+8\\b_{n+1}=\frac12n+7\frac12\\\\b_{3n}=3n-\frac12*3n+8\\b_{3n}=3n-\frac32n+8\\b_{3n}=\frac{6n}2-\frac{3n}2+8\\b_{3n}=\frac32n+8[/tex]
2.
[tex]a_{13}=0\\a_{29}=8\\\\\left \{ {{0=a_1+12r /*(-1)} \atop {8=a_1+28r}} \right. \\+\left \{ {{0=-a_1-12r \atop {8=a_1+28r}} \right. \\8=-12r+28r\\8=16r /:16\\\frac8{16}=r\\r=\frac12\\0=a_1+12*\frac12\\0=a_1+6 /-6\\-6=a_1\\\\a_n=-6+(n-1)r[/tex]
3.
[tex]a_6=\frac{32}{27}\\q=-\frac32\\a_n=a_1*q^{n-1}\\a_6=a_1*q^{5}\\\frac{32}{27}=a_1*(-\frac32)^5\\\frac{32}{27}=a_1*(-\frac{243}{32}}) /*(-\frac{32}{243})\\\frac{32}{27}*(-\frac{32}{243})=a_1\\a_1=-\frac{1024}{6561}[/tex]