Chapter 12.3, Problem 38AYU

To determine

To find: The $n\text{th}$ term of the geometric sequence.

Answer to Problem 38AYU

The $n\text{th}$ term is $a_n=\left(3\right){\left(\frac{1}{3}\right)}^{n-1}$ .

Explanation of Solution

Given information:

  $a_6=\frac{1}{81}$ and $a_3=\frac{1}{3}$ .

Calculation:

The first term of a sequence is denoted as $a_1$ and the second term of the sequence is denoted by $a_2$ .

The formula for nth term of geometric sequence is given by,

  $a_n=a_1{r}^{n-1}$ ….. (1)

Substitute $n=3$ and $a_3=\frac{1}{3}$ in above equation.

  $\begin{array}{l}\frac{1}{3}=a_1{r}^{3-1}\\ \frac{1}{3}=a_1{r}^2\end{array}$ …… (2)

Substitute $n=6$ and $a_6=\frac{1}{81}$ in above equation.

  $\begin{array}{l}\frac{1}{81}=a_1{r}^{6-1}\\ \frac{1}{81}=a_1{r}^5\end{array}$ ….. (3)

Divide equation (2) by equation (3).

  $\begin{array}{l}\frac{\frac{1}{3}}{\frac{1}{81}}=\frac{a_1{r}^2}{a_1{r}^5}\\ 27=\frac{1}{r^3}\\ r=\frac{1}{3}\end{array}$

Substitute $r=\frac{1}{3}$ in equation (2).

  $\begin{array}{l}\frac{1}{3}=a_1{\left(\frac{1}{3}\right)}^2\\ a_1=\frac{\frac{1}{3}}{\frac{1}{9}}\\ a_1=3\end{array}$

Calculate the $n\text{th}$ term.

  $a_n=\left(3\right){\left(\frac{1}{3}\right)}^{n-1}$

Therefore, the $n\text{th}$ term of the sequence is $a_n=\left(3\right){\left(\frac{1}{3}\right)}^{n-1}$ .