Chapter 3, Problem 11PFA

To determine

To find:First three terms of the sequence.

Answer to Problem 11PFA

First three terms of the sequence are $3,-1,-5$

Explanation of Solution

Given information:

  $f\left(n\right)=7-4n$

  $f\left(n\right)=7-4n$

Substitute $n=1$

We get

  $\begin{array}{l}f\left(n\right)=7-4n\\ \Rightarrow f\left(1\right)=7-4\left(1\right)\\ \Rightarrow f\left(1\right)=7-4\\ \Rightarrow f\left(1\right)=3\end{array}$

Therefore, first term of the sequence = $f\left(1\right)=3$ .

Substitute $n=2$

We get

  $\begin{array}{l}\Rightarrow f\left(2\right)=7-4\left(2\right)\\ \Rightarrow f\left(2\right)=7-8\\ \Rightarrow f\left(2\right)=-1\end{array}$

Therefore, Second term of the sequence = $f\left(2\right)=-1$

Substitute $n=3$

We get

  $\begin{array}{l}\Rightarrow f\left(3\right)=7-4\left(3\right)\\ \Rightarrow f\left(3\right)=7-12\\ \Rightarrow f\left(3\right)=-5\end{array}$

Therefore, Third term of the sequence = $f\left(3\right)=-5$

Therefore, first three terms of the sequence are $3,-1,-5$ .