Question

Find the ninth term of the sequence if the second term is 12 and the fourth term is 16/3.

Answer 1

with the assumption this is a geometric sequence and thus it as an "r" common ratio, so we know that the 3rd term must be 12 * r and the 4th term must be 12 * r * r, so let's make a quick table

$\begin{array}{rll} term&value\\ \cline{1-2} a_2&12\\ a_3&12\cdot r\\\cline{1-2} a_4&(12\cdot r)r\\ &12r^2\\ &\frac{16}{3}\\\cline{1-2} a_5&12r^3\\ a_6&12r^4\\ a_7&12r^5\\ a_8&12r^6\\ a_9&12r^7 \end{array}\qquad \implies \begin{array}{llll} 12r^2=\cfrac{16}{3}\implies r^2=\cfrac{16}{12\cdot 3}\implies r^2=\cfrac{4}{9}\\\\\\ r=\sqrt{\cfrac{4}{9}}\implies r=\cfrac{\sqrt{4}}{\sqrt{9}}\implies r=\cfrac{2}{3} \end{array} \\\\[-0.35em] ~\dotfill$

$a_9=12\left( \cfrac{2}{3} \right)^7\implies a_9=12\cdot \cfrac{128}{2187}\implies a_9=\cfrac{512}{729}$