Show that the sequence
Question:

Show that the sequence $<a_{n}>$, defined by $a_{n}=\frac{2}{3^{n}}, n \in N$ is a G.P.

Solution:

We have:

$a_{n}=\frac{2}{3^{n}}, n \in N$

Putting $n=1,2,3, \ldots$

$a_{1}=\frac{2}{3^{1}}=\frac{2}{3}, a_{2}=\frac{2}{3^{2}}=\frac{2}{9}, a_{3}=\frac{2}{3^{3}}=\frac{2}{27}$ and so on.

Now, $\frac{a_{2}}{a_{1}}=\frac{\frac{2}{9}}{\frac{9}{3}}=\frac{1}{3}, \frac{a_{3}}{a_{2}}=\frac{\frac{2}{27}}{\frac{2}{9}}=\frac{1}{3}$ and so on.

$\therefore \frac{a_{2}}{a_{1}}=\frac{a_{3}}{a_{2}}=\ldots=\frac{1}{3}$

So, the sequence is an G.P., where $\frac{2}{3}$ is the first term and $\frac{1}{3}$ is the common ratio.

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