Answer:
[tex]\large\boxed{\dfrac{1}{243}}[/tex]
Step-by-step explanation:
It's a geometric sequence:
[tex]a_1=81\\\\a_2=a_1\cdot\dfrac{1}{3}\to a_2=81\cdot\dfrac{1}{3}=27\\\\a_3=a_2\cdot\dfrac{1}{3}\to a_3=27\cdot\dfrac{1}{3}=9\\\vdots[/tex]
The explicit formula of a geometric sequence:
[tex]a_n=a_1r^{n-1}[/tex]
Substitute:
[tex]a_1=81,\ r=\dfrac{1}{3}\\\\a_n=81\left(\dfrac{1}{3}\right)^{n-1}[/tex]
Calculate the 10th term. Substitute n = 10:
[tex]a_{10}=81\left(\dfrac{1}{3}\right)^{10-1}=3^4\left(\dfrac{1}{3}\right)^9=3^4\cdot\dfrac{1}{3^9}=\dfrac{3^4}{3^9}=\dfrac{1}{3^5}=\dfrac{1}{243}[/tex]
