Question

What is the tenth term of the geometric sequence that has a common ratio of `1/3` and 36 as its fifth term?

Answer 1

$\bf \begin{array}{llll} term&value\\ \cline{1-2} a_5&36\\ a_6&36\left( \frac{1}{3} \right)\\ a_7&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\\ a_8&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\\ a_9&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\\ a_{10}&36\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right)\left( \frac{1}{3} \right) \end{array}$

$\bf a_{10}=36\left( \frac{1}{3} \right)^5\implies a_{10}=36\cdot \cfrac{1^5}{3^5}\implies a_{10}=\cfrac{36}{243}\implies a_{10}=\cfrac{4}{27}$