Chapter 11.3, Problem 28PPS

To determine

Write an equation for the nth term of the geometric series $12,-16,\frac{64}{3},\mathrm{.......}$

Answer to Problem 28PPS

  $a_n=12·{(-\frac{4}{3})}^{n-1}$

Explanation of Solution

Given: The series $12,-16,\frac{64}{3},\mathrm{.......}$

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:

  $a_1=12;\text{Common}\text{ratio}=r=\frac{\text{2nd}\text{term}}{\text{1st}\text{term}}=\frac{-16}{12}=-\frac{4}{3}$

  $\begin{array}{l}{a}_1=8,r=\frac{1}{4},\text{Find}{a}_n\\ $ a_n=a\cdot r^{n-1}=12\cdot {\left(-\frac{4}{3}\right)}^{n-1}$$ a_n=12\cdot {\left(-\frac{4}{3}\right)}^{k-1}$\end{array}$

Thus, nth term of the series $12,-16,\frac{64}{3},\mathrm{.......}$ is $a_n=12·{(-\frac{4}{3})}^{n-1}$