# How do you find the Nth term in the geometric sequence where the first term is 3 and the fourth term is 6sqrt2?

Dec 31, 2015

${a}_{N} = 3 \cdot {2}^{\frac{N - 1}{2}}$

#### Explanation:

The general term of a geometric sequence can be written:

${a}^{n} = a {r}^{n - 1}$

where $a$ is the initial term and $r$ the common ratio.

In our case we are given ${a}_{1} = 3$ and ${a}_{4} = 6 \sqrt{2}$

So we find:

${r}^{3} = \frac{a {r}^{3}}{a {r}^{0}} = {a}_{4} / {a}_{1} = \frac{6 \sqrt{2}}{3} = 2 \sqrt{2} = {\left(\sqrt{2}\right)}^{3}$

So (assuming the geometric sequence is Real):

$r = \sqrt[3]{{\sqrt{2}}^{3}} = \sqrt{2}$

and

${a}_{n} = 3 \cdot {\left(\sqrt{2}\right)}^{n - 1} = 3 \cdot {2}^{\frac{n - 1}{2}}$