Chapter 11.1, Problem 8OE

To determine

To calculate:

The common ratio and missing terms for the following arithmetic sequence:

  $?,?,\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81}$

Answer to Problem 8OE

The common ratio is $25$ and missing terms are $3$ and $1$ .

Explanation of Solution

Given information:

  $?,?,\frac{1}{3},\frac{1}{9},\frac{1}{27},\frac{1}{81}$

Calculation:

Consider the sequence as:

  $\begin{array}{l}{a}_1=?\\ a_2=?\\ a_3=\frac{1}{3}\\ a_4=\frac{1}{9}\\ a_5=\frac{1}{27}\\ a_6=\frac{1}{81}\end{array}$

Now, the common ratio is calculated as:

  $\begin{array}{l}r=\frac{a_4}{a_3}=\frac{\left(\frac{1}{9}\right)}{\left(\frac{1}{3}\right)}=\frac{3}{9}=\frac{1}{3}\\ r=\frac{a_5}{a_4}=\frac{\left(\frac{1}{27}\right)}{\left(\frac{1}{9}\right)}=\frac{9}{27}=\frac{1}{3}\\ r=\frac{a_6}{a_5}=\frac{\left(\frac{1}{81}\right)}{\left(\frac{1}{27}\right)}=\frac{27}{81}=\frac{1}{3}\end{array}$

The first is calculated by using the last term as:

  $\begin{array}{l}{a}_6=a_1{r}^{6-1}\\ \frac{1}{81}=a_1{\left(\frac{1}{3}\right)}^5\\ a_1=\frac{1}{81}\times 243\\ a_1=3\end{array}$

Now, the second terms is calculated as:

  $\begin{array}{l}{a}_2=a_1{r}^{2-1}\\ a_2=3{\left(\frac{1}{3}\right)}^1\\ a_2=1\end{array}$