Chapter 12, Problem 20RE

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 20RE

The series is arithmetic, common difference is $-4$ and the sum of first $n$ term is $n\left(3-2n\right)$ .

Explanation of Solution

Given: $1,-3,-7,-11,\mathrm{...}$

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

  $\begin{array}{c}{b}_2-b_1=b_3-b_2\\ -3-1=-7-\left(-3\right)\\ -4=-7+3\\ -4=-4\end{array}$

So, the common difference is $-4$ .

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

  $\begin{array}{c}{S}_n=\frac{n}{2}\left(2a_1+\left(n-1\right)d\right)\\ =\frac{n}{2}\left(2\cdot 1+\left(n-1\right)-4\right)\\ =\frac{n}{2}\left(2-4n+4\right)\\ =n\left(3-2n\right)\end{array}$

Therefore, the series is arithmetic, common difference is $-4$ and the sum of first $n$ term is $n\left(3-2n\right)$