Chapter 8.3, Problem 29E

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 29E

First five terms of the geometric sequence are $6,-9,\frac{27}{2},-\frac{81}{4},\frac{243}{8}$ respectively.

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

An expression for nth term of the sequence is $a_n=6{\left(-\frac{3}{2}\right)}^{n-1}$ .

Explanation of Solution

Given information:

An sequence is given with first term $a_1=6$ and a recurrence relation $a_{k+1}=-\frac{3}{2}{a}_k$ .

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.

  $a_1=6$

And

  $a_{k+1}=-\frac{3}{2}{a}_k$

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

  $\begin{array}{l}{a}_2=-\frac{3}{2}{a}_1=-\frac{3}{2}\times 6=-9\\ a_3=-\frac{3}{2}{a}_2=-\frac{3}{2}\times \left(-9\right)=\frac{27}{2}\\ a_4=-\frac{3}{2}{a}_3=-\frac{3}{2}\times \frac{27}{2}=-\frac{81}{4}\\ a_5=-\frac{3}{2}{a}_4=-\frac{3}{2}\times \left(-\frac{81}{4}\right)=\frac{243}{8}\end{array}$

Hence, the first five terms of the sequence are $6,-9,\frac{27}{2},-\frac{81}{4},\frac{243}{8}$ respectively.

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

  $\begin{array}{c}r=\frac{a_2}{a_1}\\ =\frac{-9}{6}\\ =-\frac{3}{2}\end{array}$

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

Now, the expression for nth term can be found as

  $\begin{array}{c}{a}_n=a_1{r}^{n-1}\\ =6{\left(-\frac{3}{2}\right)}^{n-1}\\ a_n=6{\left(-\frac{3}{2}\right)}^{n-1}\end{array}$

Hence, the expression for nth term of the sequence is $a_n=6{\left(-\frac{3}{2}\right)}^{n-1}$ .