Chapter 11.3, Problem 34PPS

To determine

Write an equation for the nth term of the geometric series with the information $a_4=80;r=4$

Answer to Problem 34PPS

  $a_n=5·4^{n-2}$

Explanation of Solution

Given:

The information:$a_4=80;r=4$

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:

Use the Exponent rules:$\text{Product}\text{rule}=a^m·a^n=a^{m+n}$

  $\text{Negative}\text{exponent}\text{rule}:a^{-n}=\frac{1}{a^n}or\frac{1}{a^{-n}}=a^n$

Step 1: Find the value of $a_1$ from the information $a_4=80;r=4$

  $\begin{array}{}$ a_n=a\cdot r^{n-1}$$ a_4=80r=4$$ a_4=a\cdot r^{4-1}=a_1\cdot r^3$$ 80=a_1\cdot {(4)}^3=64a_1$$ \frac{80}{64}=\frac{64}{64}a$$ a_1=\frac{5}{4}$\end{array}$

  $\begin{array}{}$ a_1=\frac{5}{4};r=4$$ a_n=a\cdot r^{n-1}=\frac{5}{4}\cdot {(4)}^{n-4}$$ a_n=5\cdot 4^{-1}\cdot {(4)}^{n-1}=5\cdot 4^{-1+n-1}=5\cdot 4^{n-2}$$ a_n=5\cdot 4^{n-2}$\end{array}$

Thus, nth term of the series is $a_n=5·4^{n-2}$