Question

Find the 8th term of the geometric sequence in which a3=4/3 and a6=36

Answer 1

`\begin{array}{rll} term&value\\ \cline{1-2} a_3&(4)/(3)\\\\ a_4&(4)/(3)\cdot r\\\\ a_5&(4)/(3)\cdot r\cdot r\\\\ a_6&(4)/(3)\cdot r\cdot r\cdot r\\\\ &(4)/(3)\cdot r^3~~ = ~~36 \end{array}\hspace{7em} \begin{array}{llll} \cfrac{4}{3}r^3=36\implies r^3=36\cdot \cfrac{3}{4} \\\\\\ r^3=27\implies r=\sqrt[3]{27}\stackrel{ \textit{common ratio} }{\implies r=3} \end{array} \\\\[-0.35em] ~\dotfill\\\\ a_7=36(3)\hspace{5em}a_8=36(3)(3)\implies a_8=324`