Problems from General term of a geometric progression

Find the terms that are in the fourth and eighth position in the geometric progression with reason $r = 0, 3$ and the first term $a_{1} = 1, 25.$

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We know that the general term of our progression is:

$a_{n} = 1, 25 \cdot (0, 3)^{n - 1} = \frac{5}{4} \cdot (\frac{3}{10})^{n - 1}$

So the terms we are looking for are:

$a_{4} = \frac{5}{4} \cdot (\frac{3}{10})^{3} = \frac{5 \cdot 27}{4 \cdot 1.000} = \frac{135}{4.000} = 0, 03375$

$a_{8} = \frac{5}{4} \cdot (\frac{3}{10})^{7} = \frac{5 \cdot 2.187}{4 \cdot 10.000 .000} = \frac{10.935}{40.000 .000} = 0, 000273$

$a_{4} = 3, 37 \cdot 10^{- 2}$ and $a_{8} = 2, 73 \cdot 10^{- 4} .$

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Find the general term of the geometric progression $(2, 2 \sqrt[3]{3}, 2 \sqrt[3]{9}, 6, 6 \sqrt[3]{3}, \dots)$

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

$r = \frac{a_{n}}{a_{n - 1}} = \frac{a_{2}}{a_{1}} = \frac{2 \sqrt[3]{3}}{2} = \sqrt[3]{3}$

And, knowing that the first term is $a_{1} = 2$ , we have:

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

$a_{n} = 2 \sqrt[3]{3^{n - 1}}$

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