Chapter 11.3, Problem 34WE

To determine

To calculate:

The following sequence is arithmetic, geometric and the formula for $n^{th}$ term:

  $\frac{a^2}{2},\frac{a^4}{4},\frac{a^6}{8}$

Answer to Problem 34WE

The sequence is geometric sequence and formula for $n^{th}$ term is $t_n={\left(\frac{a^2}{2}\right)}^n$ .

Explanation of Solution

Given information:

  $\frac{a^2}{2},\frac{a^4}{4},\frac{a^6}{8}$

Calculation:

Check the common difference first to categorize the given sequence into arithmetic, geometric or neither:

  $\begin{array}{l}{a}_1=\frac{a^2}{2}\\ a_2=\frac{a^4}{4}\\ a_3=\frac{a^6}{8}\end{array}$

Now, the common difference is calculated as:

  $\begin{array}{l}d=\frac{a_2}{a_1}=\frac{\left(\frac{a^4}{4}\right)}{\left(\frac{a^2}{2}\right)}=\frac{2a^4}{4a^2}=\frac{a^2}{2}\\ d=\frac{a_3}{a_2}=\frac{\left(\frac{a^6}{8}\right)}{\left(\frac{a^4}{4}\right)}=\frac{4a^6}{8a^4}=\frac{a^2}{2}\end{array}$

Therefore, the given sequence is geometric sequence as the common ratio is same is $\frac{a^2}{2}$ .

Now, the formula for $n^{th}$ term is calculated as:

  $\begin{array}{l}{t}_n=a_1{r}^{n-1}\\ t_n=\left(\frac{a^2}{2}\right){\left(\frac{a^2}{2}\right)}^{n-1}\\ t_n={\left(\frac{a^2}{2}\right)}^{1+n-1}\\ t_n={\left(\frac{a^2}{2}\right)}^n\end{array}$