Chapter 9.3, Problem 29E

To determine

To write: an expression for the $n$ th term of the geometric sequence and find the ${12}^{th}$ term.

Answer to Problem 29E

The expression for nth term is $a_n={\left(2\right)}^{\frac{n-1}{2}}$ .

The ${12}^{th}$ term of sequence is $a_{12}=32\sqrt{2}$ .

Explanation of Solution

Given information:

The given first term and common ratio,

  $a_1=1,r=\sqrt{2}$

Calculation:

Consider the first time $a_1=1$ and common ration $r=\sqrt{2}$

The general term of geometric sequence is given by formula

  $a_n=a_1{r}^{n-1}$

Here, $a_1$ is the first term and $r$ is common ratio.

Substitute $a_1=1$ and $r=\sqrt{2}$ as shown below.

  $\begin{array}{c}{a}_n=\left(1\right){\left(\sqrt{2}\right)}^{n-1}\\ a_n={\left(2^{\frac{1}{2}}\right)}^{n-1}\\ a_n={\left(2\right)}^{\frac{n-1}{2}}\end{array}$

Therefore, the formula for general term of geometric sequence is $a_n={\left(2\right)}^{\frac{n-1}{2}}$ .

Substitute $n=12$ into the formula $a_n={\left(2\right)}^{\frac{n-1}{2}}$ and simplify as shown below.

  $\begin{array}{c}{a}_{12}={\left(2\right)}^{\frac{n-1}{2}}\\ ={\left(2\right)}^{\frac{12-1}{2}}\\ =2^{\frac{11}{2}}\\ =32\sqrt{2}\end{array}$

Therefore, the ${12}^{th}$ term of sequence is $a_{12}=32\sqrt{2}$ .