Chapter 12.3, Problem 40AYU

In Problems 33-40, find the $n^{\text{th}}$ term $a_n$ of each geometric sequence. When given, $r$ is the common ratio.

$a_3=\frac{1}{3};a_6=\frac{1}{81}$

To determine

To calculate: The value of the $n^{th}$ term of the geometric sequence using the third and sixth term values.

Answer to Problem 40AYU

Solution:

The $n^{th}$ term of a geometric sequence using the given data is $a_n=3\cdot {\left(\frac{1}{3}\right)}^{n-1}$.

Explanation of Solution

Given:

The third term value, $a_3=\frac{1}{3}$ and the sixth term value, $a_6=\frac{1}{81}$.

Formula Used:

The nth term of the geometric sequence is given by,

$a_n=a_1\cdot r^{n-1}$

Consider the third term, $\frac{1}{3}=a_3=a_1\cdot r^{3-1}=a_1{r}^2$   -----(1)

Now consider the sixth term , $\frac{1}{81}=a_6=a_1\cdot r^{6-1}=a_1{r}^5$   -----(2)

Divide equation (2) by equation (1), $\frac{a_1{r}^5}{a_1{r}^2}=\frac{\frac{1}{81}}{\frac{1}{3}}=\frac{1}{27}$.

$r^3=\frac{1}{27}$

$r=\frac{1}{3}$

From (1), $a_1{r}^2=\frac{1}{3}$.

$a_1{\left(\frac{1}{3}\right)}^2=\frac{1}{3}$

$a_1=3$

Substituting the value of $a_1$ and the common ratio $r$ in the $n^{\text{th}}$ term,

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