General term of a geometric progression

Let's find the quotient of two consecutive terms to find the ratio:

$$r=\dfrac{a_n}{a_{n-1}}=\dfrac{a_2}{a_1}=\dfrac{2\sqrt[3]{3}}{2}=\sqrt[3]{3}$$

And, knowing that the first term is $a_1=2$, we have:

$$a_n=a_1 \cdot r^{n-1}=2(\sqrt[3]{3})^{n-1}=2\sqrt[3]{3^{n-1}}=2\cdot 3^{\frac{n-1}{3}}$$

We know that the general term of our progression is:

$$a_n=1,25\cdot (0,3)^{n-1}=\dfrac{5}{4}\cdot \Big(\dfrac{3}{10}\Big)^{n-1}$$

So the terms we are looking for are:

$$a_4=\dfrac{5}{4}\cdot \Big(\dfrac{3}{10}\Big)^{3}=\dfrac{5\cdot27}{4\cdot 1.000}=\dfrac{135}{4.000}=0,03375$$

$$a_8=\dfrac{5}{4}\cdot \Big(\dfrac{3}{10}\Big)^{7}=\dfrac{5\cdot2.187}{4\cdot 10.000.000}=\dfrac{10.935}{40.000.000}=0,000273$$