For the following geometric sequence

For the following geometric sequence, find a 10.

1, 1/3, 1/9, 1/27

$a_1=1 \\ a_2=\frac13 \\ a_3=\frac19 \\ a_4=\frac{1}{27} \\ \boxed{~ r = \frac{a_{n}}{a_{n-1}} ~}\\ r=\frac{ a_4 }{ a_3 }=\frac{ a_3 }{ a_2 }=\frac{ a_2 }{ a_1 }\\ r=\frac{ \frac{1}{27}}{ \frac19 }=\frac{ \frac19 }{ \frac13 }=\frac{ \frac13 }{ 1 }\\ r=\frac{9}{27}=\frac{ 3 }{9 }=\frac13\\ r=\frac{1}{3}=\frac{ 1 }{3}=\frac13\\ \mathbf{r=\frac13}$

$\boxed{~ a_n = a_1\cdot r^{n-1} ~}\\ a_1 = 1\cdot \left( \frac13 \right)^{1-1} = 1\\ a_2 = 1\cdot \left( \frac13 \right)^{2-1} = 1\cdot \frac13= \frac13\\ a_3 = 1\cdot \left( \frac13 \right)^{3-1} = 1\cdot \left( \frac13 \right)^2 = \frac19\\ a_4 = 1\cdot \left( \frac13 \right)^{4-1} = 1\cdot \left( \frac13 \right)^3 = \frac{1}{27}\\ \dots \\ a_{10} = 1\cdot \left( \frac13 \right)^{10-1} = 1\cdot \left( \frac13 \right)^9 = \frac{1}{19683}\\ \mathbf{a_{10} = \frac{1}{19683}}\\$

Notice that each successive term is 1/3 as large as the previous one.....so.....the common ratio = 1/3

And the "formula" for the nth term is just  (1/3)^n-1

So....the tenth term is :  (1/3)^{(10 - 1)}  = (1/3)⁹ = 1 / 19683