Answer: The 10-th term of the sequence is [tex]\dfrac{4}{27}.[/tex]
Step-by-step explanation: We are to find the 10-th term of a geometric sequence that has common ratio [tex]\dfrac{1}{3}[/tex] and fifth term is 36.
We know that the n-th term of a geometric sequence with first-term 'a' and common ratio 'r' is given by
[tex]a_n=a r^{n-1}.[/tex]
According to the given information, we have
[tex]r=\dfrac{1}{3},[/tex]
and
[tex]a_5=36\\\\\Rightarrow a r^{5-1}=36\\\\\Rightarrow ar^4=36\\\\\Rightarrow a\times \left(\dfrac{1}{3}\right)^4=36\\\\\Rightarrow a=36\times 81\\\\\Rightarrow a=2916.[/tex]
Therefore, the 10-th term of the sequence will be
[tex]a_{10}=a r^{10-1}=2916\times r^9=2916\times \left(\dfrac{1}{3}\right)9\\\\\\\Rightarrow a_{10}=\dfrac{3^6\times 4}{3^9}\\\\\\\Rightarrow a_{10}=\dfrac{4}{27}.[/tex]
Thus, the 10-th term of the sequence is [tex]\dfrac{4}{27}.[/tex]
