Chapter 9.3, 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 $4,\frac{4x}{5},\frac{4x^2}{25},\frac{4x^3}{125},\frac{4x4}{625}$ .

Explanation of Solution

Given information:

The given sequence,

  $a_1=4,r=x/5$

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:

Consider the first term $a_1=4$ and common ratio $r=\frac{x}{5}$ :

The first term is $a_1=4$ .

To find the second term, multiply first term by $\frac{x}{5}$ as shown blow

  $\begin{array}{c}{a}_2=a_{1}r\\ =4\cdot \frac{x}{5}\\ =\frac{4x}{5}\end{array}$

So, the second term is $\frac{4x}{5}$ .

The 3^rd term can be calculated as

  $\begin{array}{c}{a}_3=a_{2}r\\ =\frac{4x}{5}\cdot \frac{x}{5}\\ =\frac{4x^2}{25}\end{array}$

So, the third term is $\frac{4x^2}{25}$

The 4^th term can be calculated as

  $\begin{array}{c}{a}_4=a_{3}r\\ =\frac{4x^2}{25}\cdot \frac{x}{5}\\ =\frac{4x^3}{125}\end{array}$

So, the fourth term is $\frac{4x^3}{125}$ .

The 5^th term can be calculated as

  $\begin{array}{c}{a}_5=a_{4}r\\ =\frac{4x^3}{125}\cdot \frac{x}{5}\\ =\frac{4x^4}{625}\end{array}$

So, the fifth term is $\frac{4x^4}{625}$

Therefore, the five terms of sequence are $4,\frac{4x}{5},\frac{4x^2}{25},\frac{4x^3}{125},\frac{4x4}{625}$ .