Find the 12th term and the sum of the first 12 terms of a geometric sequence. A geometric series has a first term $\sqrt{2}$ and a second term $\sqrt{6}$ . Find the 12th term and the sum of the first 12 terms.
I can get to the answers as irrational numbers using a calculator but how can I can obtain the two answers in radical form $243 * \sqrt{6}$ and $364 \left(\sqrt{6}+\sqrt{2}\right)$ ?
The closest I get with the 12th term is $\sqrt{2} \left(\sqrt{6} \over \sqrt{2}\right)^{(12-1)}$  or  $\sqrt{2} * 3^\left({11\over 2}\right)$
And for the sum ${\sqrt{2}-\sqrt{2}*(\sqrt{3})^{12} \over 1 - \sqrt{3}}$

So, the common ratio $=\frac{\sqrt6}{\sqrt2}=\sqrt3$
So, the $n$ th term $=\sqrt2(\sqrt3)^{n-1}\implies 12$th term $=\sqrt2(\sqrt3)^{12-1}=\sqrt2(\sqrt3)^{11}$
Now, $\displaystyle(\sqrt3)^{11}=\sqrt3 \cdot 3^5=243\sqrt3$
The sum of $n$ term is $\displaystyle \sqrt2\cdot\frac{(\sqrt3)^n-1}{\sqrt3-1}$
$\implies 12$th term $=\displaystyle \sqrt2\cdot\frac{(\sqrt3)^{12}-1}{\sqrt3-1}=\sqrt2\cdot\frac{(3^6-1)(\sqrt3+1)}{(\sqrt3-1)(\sqrt3+1)}$ (rationalizing the denominator )
$\displaystyle=\frac{(3^3-1)(3^3+1)\sqrt2(\sqrt3+1)}2=364(\sqrt6+\sqrt2)$
