Given: the geometric sequence whose common ratio is 1/3 and whose first term is 6.
Find: 9th term of the geometric sequence
Explanation:
[tex]\begin{gathered} a_n=a_1(r)^{n-1} \\ a_9=6(\frac{1}{3})^{9-1} \\ a_9=6(\frac{1}{3})^8 \end{gathered}[/tex]Final answer: the required answer is
[tex]\frac{6}{3^8}[/tex]