# 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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### Development:

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

$$a_8=\dfrac{5}{4}\cdot \Big(\dfrac{3}{10}\Big)^{7}=\dfrac{5\cdot2.187}{4\cdot 10.000.000}=\dfrac{10.935}{40.000.000}=0,000273$$$### Solution: $$a_4=3,37\cdot 10^{-2}$$ and $$a_8=2,73\cdot 10^{-4}.$$ Hide solution and development Find the general term of the geometric progression $$(2, 2\sqrt[3]{3}, 2\sqrt[3]{9}, 6, 6\sqrt[3]{3}, \ldots)$$$

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### Development:

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

### Solution:

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

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