Chapter 8.3, Problem 26E

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 26E

First five terms of the geometric sequence are $81,27,9,3,1$ respectively.

Common ratio of the sequence is $\frac{1}{3}$ .

An expression for nth term of the sequence is $a_n=81{\left(\frac{1}{3}\right)}^{n-1}$ .

Explanation of Solution

Given information:

An sequence is given with first term $a_1=81$ and a recurrence relation $a_{k+1}=\frac{1}{3}{a}_k$ .

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.

  $a_1=81$

And

  $a_{k+1}=\frac{1}{3}{a}_k$

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

  $\begin{array}{l}{a}_2=\frac{1}{3}{a}_1=\frac{1}{3}\times 81=27\\ a_3=\frac{1}{3}{a}_2=\frac{1}{3}\times 27=9\\ a_4=\frac{1}{3}{a}_3=\frac{1}{3}\times 9=3\\ a_5=\frac{1}{3}{a}_4=\frac{1}{3}\times 3=1\end{array}$

Hence, the first five terms of the sequence are $81,27,9,3,1$ respectively.

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

  $\begin{array}{c}r=\frac{a_2}{a_1}\\ =\frac{27}{81}\\ =\frac{1}{3}\end{array}$

Therefore, the common ratio is $\frac{1}{3}$ .

Now, the expression for nth term can be found as

  $\begin{array}{c}{a}_n=a_1{r}^{n-1}\\ =81{\left(\frac{1}{3}\right)}^{n-1}\\ a_n=81{\left(\frac{1}{3}\right)}^{n-1}\end{array}$

Hence, the expression for nth term of the sequence is $a_n=81{\left(\frac{1}{3}\right)}^{n-1}$ .