Chapter 12, Problem 18RE

To determine

To find:Whether the series is arithmetic, geometric or neither, if the series is arithmetic find the common difference and sum of first $n$ terms and if the series is geometric find the common ratio and the sum of first $n$ terms.

Answer to Problem 18RE

The sequence is geometric, common ratio is $9$ and the sum of the $n$ terms is $\frac{9}{8}\left(1-9^n\right)$ .

Explanation of Solution

Given: $\left\{u_n\right\}=\left\{3^{2n}\right\}$

The terms of the series are,

  $\begin{array}{l}{u}_1=3^{2\left(1\right)}=9\\ u_2=3^{2\left(2\right)}=81\\ u_3=3^{2\left(3\right)}=729\\ u_4=3^{2\left(4\right)}=6567\end{array}$

If the series is arithmetic then it must the follow the relation,.

  $\begin{array}{c}{u}_2-u_1=u_3-u_2\\ 81-9\ne 729-81\end{array}$

The sequence is not arithmetic.

Foe the sequence to be geometric it must follow,

  $\begin{array}{c}\frac{u_2}{u_1}=\frac{u_3}{u_2}\\ \frac{81}{9}=\frac{729}{81}\\ 9=9\end{array}$

So, the sequence is geometric.

The common ratio is $9$ .

The sum is calculated as,’

  $\begin{array}{c}T=9\left(\frac{1-9^n}{1-9}\right)\\ =\frac{9}{8}\left(1-9^n\right)\end{array}$

Therefore, the sequence is geometric, common ratio is $9$ and the sum of the $n$ terms is $\frac{9}{8}\left(1-9^n\right)$