Chapter 8.3, Problem 30E

To determine

To find: first five terms of the geometric sequence and the common ratio. Write the nth term of the sequence as a function of n .

Answer to Problem 30E

First five terms of the geometric sequence are $30,-20,\frac{40}{3},-\frac{80}{9},\frac{160}{18}$ respectively.

Common ratio of the sequence is $-\frac{2}{3}$ .

An expression for nth term of the sequence is $a_n=30{\left(-\frac{2}{3}\right)}^{n-1}$ .

Explanation of Solution

Given information:

An sequence is given with first term $a_1=30$ and a recurrence relation $a_{k+1}=-\frac{2}{3}{a}_k$ .

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:

Now, consider the first term.

  $a_1=30$

And

  $a_{k+1}=-\frac{2}{3}{a}_k$

First five terms of the sequence can be found by substituting 1, 2, 3, 4, and 5 for kas shown:

  $\begin{array}{l}{a}_2=-\frac{2}{3}{a}_1=-\frac{2}{3}\times 30=-20\\ a_3=-\frac{2}{3}{a}_2=-\frac{2}{3}\times \left(-20\right)=\frac{40}{3}\\ a_4=-\frac{2}{3}{a}_3=-\frac{2}{3}\times \left(\frac{40}{3}\right)=-\frac{80}{9}\\ a_5=-\frac{2}{3}{a}_4=-\frac{2}{3}\times \left(-\frac{80}{9}\right)=\frac{160}{18}\end{array}$

Hence, the first five terms of the sequence are $30,-20,\frac{40}{3},-\frac{80}{9},\frac{160}{18}$ respectively.

The common ratio can be found by dividing first term by second term.

  $\begin{array}{c}r=\frac{a_2}{a_1}\\ =\frac{-20}{30}\\ =-\frac{2}{3}\end{array}$

Therefore, the common ratio is $-\frac{2}{3}$ .

Now, the expression for nth term can be found as

  $\begin{array}{c}{a}_n=a_1{r}^{n-1}\\ =30{\left(-\frac{2}{3}\right)}^{n-1}\\ a_n=30{\left(-\frac{2}{3}\right)}^{n-1}\end{array}$

Hence, the expression for nth term of the sequence is $a_n=30{\left(-\frac{2}{3}\right)}^{n-1}$ .