Find the Nth term in the geometric sequence where the first term is 3 and the fourth term is 6√2

Find the Nth term in the geometric sequence where the first term is 3 and the fourth term is $6\sqrt{2}$
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Szulikto
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:
${r}^{3}=\frac{a{r}^{3}}{a{r}^{0}}=\frac{{a}_{4}}{{a}_{1}}=\frac{6\sqrt{2}}{3}=2\sqrt{2}={\left(\sqrt{2}\right)}^{3}$
So (assuming the geometric sequence is Real):
$r=\sqrt[3]{{\sqrt{2}}^{3}}=\sqrt{2}$
and
${a}_{n}=3\cdot {\left(\sqrt{2}\right)}^{n-1}=3\cdot {2}^{\frac{n-1}{2}}$