# Question 6f584

Jan 13, 2018

${2}^{\frac{11}{2}}$

#### Explanation:

The $n$th term of the series is given by ${a}_{n} = 2 \cdot {\left(\sqrt{2}\right)}^{n - 1}$.

The tenth term is ${a}_{10} = 2 {\left(\sqrt{2}\right)}^{9} = 2 \cdot {\left({2}^{\frac{1}{2}}\right)}^{9}$

$= 2 \cdot {2}^{\frac{9}{2}} = {2}^{\frac{11}{2}}$

Jan 13, 2018

${a}_{10} = 32 \sqrt{2}$

#### Explanation:

$\text{the n th term of a geometric sequence is }$

•color(white)(x)a_n=ar^(n-1)#

$\text{where a is the first term and r the common ratio}$

$\text{here "a=2" and } r = \sqrt{2}$

$\Rightarrow {a}_{10} = 2 \times {\left(\sqrt{2}\right)}^{9}$

$\left[\text{note that } {\left(\sqrt{2}\right)}^{2} = 2\right]$

$\Rightarrow {a}_{10} = 2 \times {2}^{4} \sqrt{2} = {2}^{5} \sqrt{2} = 32 \sqrt{2}$