Chapter 12, Problem 13RE

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 13RE

The series is arithmetic, common difference is $1$ and the sum of first $n$ term is $\frac{n}{2}\left(n+11\right)$ .

Explanation of Solution

Given: $\left\{a_n\right\}=\left\{n+5\right\}$

The terms of the series are,

  $\begin{array}{c}{a}_1=1+5=6\\ a_2=2+5=7\\ a_3=3+5=8\\ a_4=4+5=9\end{array}$

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

  $\begin{array}{c}{a}_2-a_1=a_3-a_2\\ 7-6=8-7\\ 1=1\end{array}$

So, the common difference is $1$ .

The sum of first $n$ terms is calculated as,

  $\begin{array}{c}{S}_n={\displaystyle \sum _{k=1}^n\left(a_1+\left(k-1\right)d\right)}\\ ={\displaystyle \sum _{k=1}^n\left(6+\left(k-1\right)1\right)}\\ ={\displaystyle \sum _{k=1}^n\left(k+5\right)}\\ ={\displaystyle \sum _{k=1}^n\left(k\right)}+{\displaystyle \sum _{k=1}^n\left(5\right)}\end{array}$

Solve further,

  $\begin{array}{c}{S}_n=\frac{n\left(n+1\right)}{2}+5n\\ =\frac{n}{2}\left(n+11\right)\end{array}$

Therefore, the series is arithmetic, common difference is $1$ and the sum of first $n$ term is $\frac{n}{2}\left(n+11\right)$ .