Answer:
[tex]a_n=3\left(\sqrt{2}\right)^{n-1}[/tex]
[tex]a_{10}=48 \sqrt{2}[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Geometric sequence}\\\\$a_n=ar^{n-1}$\\\\where:\\\phantom{ww}$\bullet$ $a$ is the first term. \\\phantom{ww}$\bullet$ $r$ is the common ratio.\\\phantom{ww}$\bullet$ $a_n$ is the $n$th term.\\\phantom{ww}$\bullet$ $n$ is the position of the term.\\\end{minipage}}[/tex]
Given:
- [tex]a = 3[/tex]
- [tex]r = \sqrt{2}[/tex]
- [tex]n=10[/tex]
Substitute the given values of a and r into the formula to create an equation for the nth term:
[tex]a_n=3\left(\sqrt{2}\right)^{n-1}[/tex]
To find the 10th term, substitute n = 10 into the equation:
[tex]\implies a_{10}=3\left(\sqrt{2}\right)^{10-1}[/tex]
[tex]\implies a_{10}=3\left(\sqrt{2}\right)^{9}[/tex]
[tex]\implies a_{10}=3 \cdot 2^{\frac{9}{2}}[/tex]
[tex]\implies a_{10}=3 \cdot 2^{4+\frac{1}{2}}[/tex]
[tex]\implies a_{10}=3 \cdot 2^{4} \cdot 2^{\frac{1}{2}}[/tex]
[tex]\implies a_{10}=3 \cdot 16 \cdot \sqrt{2}[/tex]
[tex]\implies a_{10}=48 \sqrt{2}[/tex]
