Chapter 8.3, Problem 22E

To determine

To write: the first five terms of the geometric sequence.

Answer to Problem 22E

The first five terms of the geometric sequence are $6,-\frac{3}{2},\frac{3}{8},-\frac{3}{32},\frac{3}{128}$ ..

Explanation of Solution

Given information:

The given sequence,

  $a_1=6,r=-\frac{1}{4}$

Concept used:

The $n$ th term of geometric sequence has the form $a_1=a_1{r}^{n-1}$ where $r$ is the common ratio of the consecutive terms of the sequence.

Every geometric sequence can be written in the following form.

  $a_1=a_1,a_2=a_{1}r,a_3=a_1{r}^2,a_4=a_1{r}^3,\mathrm{...},a_n=a_1{r}^{n-1},\mathrm{..}$

Calculation:

The common ratio and the first term of the sequence is given

  $\begin{array}{l}{a}_1=6\\ r=-\frac{1}{4}\end{array}$

To find the second term, multiply first term $-\frac{1}{4}$ as shown below.

  $\begin{array}{c}{a}_2=6\left(-\frac{1}{4}\right)\\ =-\frac{3}{2}\end{array}$

The 3^rd term can be calculated as

  $\begin{array}{c}{a}_3=6{\left(-\frac{1}{4}\right)}^2\\ =\frac{3}{8}\end{array}$

The 4^th term can be calculated as

  $\begin{array}{c}{a}_4=6{\left(-\frac{1}{4}\right)}^3\\ =-\frac{3}{32}\end{array}$

The 5^th term can be calculated as

  $\begin{array}{c}{a}_5=6{\left(-\frac{1}{4}\right)}^4\\ =-\frac{3}{128}\end{array}$

Hence, the first five terms of the geometric sequence are

  $6,-\frac{3}{2},\frac{3}{8},-\frac{3}{32},\frac{3}{128}$