Chapter 11.3, Problem 6CYU

To determine

Write an equation for the nth term of the geometric series with the information $a_6=\frac{1}{8};r=\frac{3}{4}$

Answer to Problem 6CYU

  $a_n=\frac{128}{243}·{(\frac{3}{4})}^{n-1}$

Explanation of Solution

Given:

The information:$a_6=\frac{1}{8};r=\frac{3}{4}$

Concept Used:

General equation for the nth term of any geometric series is $a_n=a_1·r^{n-1}$

Where $a_1$ the first term and r is the common ratio.

Calculation:

Step 1: Find the value of $a_1$ from the information $a_6=\frac{1}{8};r=\frac{3}{4}$

  $\begin{array}{l}{a}_n=a_1·r^{n-1}\\ a_6=\frac{1}{8};r=\frac{3}{4}\\ a_6=a_1·r^{6-1}=a_1·r^5\\ $ \frac{1}{8}=a\cdot {\left(\frac{3}{4}\right)}^5=\frac{243}{1024}a$\frac{1024}{243}·\frac{1}{8}=\frac{1024}{243}·\frac{243}{1024}{a}_1\\ a_1=\frac{128}{243}\end{array}$

  $\begin{array}{l}{a}_6=\frac{1}{8};r=\frac{3}{4}\\ a_1=\frac{128}{243};r=\frac{3}{4}\\ $ a=a\cdot r^{n-1}=\frac{128}{243}\cdot {\left(\frac{3}{4}\right)}^{n-1}$$ a=\frac{128}{243}\cdot {\left(\frac{3}{4}\right)}^{n-1}$\end{array}$

Thus, nth term of the series is $a_n=\frac{128}{243}·{(\frac{3}{4})}^{n-1}$