Chapter 12, Problem 16RE

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 16RE

The sequence is neither arithmetic nor geometric.

Explanation of Solution

Given: $\left\{d_n\right\}=\left\{2n^2-1\right\}$

The terms of the series are,

  $\begin{array}{l}\left\{d_1\right\}=\left\{2{\left(1\right)}^2-1\right\}=1\\ \left\{d_2\right\}=\left\{2{\left(2\right)}^2-1\right\}=7\\ \left\{d_3\right\}=\left\{2{\left(3\right)}^2-1\right\}=17\\ \left\{d_4\right\}=\left\{2{\left(4\right)}^2-1\right\}=31\end{array}$

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

  $\begin{array}{c}{d}_2-d_1=d_3-d_2\\ 7-1=17-7\\ 6\ne 10\end{array}$

The sequence is not arithmetic.

Foe the sequence to be geometric it must follow,

  $\begin{array}{c}\frac{d_2}{d_1}=\frac{d_3}{d_2}\\ \frac{7}{1}=\frac{17}{7}\\ 7\ne \frac{17}{7}\end{array}$

So, the sequence is not geometric.

Therefore, the sequence is neither arithmetic nor geometric.