Chapter 7.2, Problem 64MR

To determine

To Find: The third, seventh and tenth terms of the sequence described by the rule $A\left(n\right)=1.2+\left(n-1\right)\left(-4\right)$ .

Answer to Problem 64MR

The required third, seventh and tenth term of the sequence is -6.8, -22.8 and -34.8 respectively.

Explanation of Solution

Given information:

The sequence $A\left(n\right)=1.2+\left(n-1\right)\left(-4\right)$

Calculation:

  $A\left(n\right)=1.2+\left(n-1\right)\left(-4\right)$

To find the third term, substitute the term number as 3 i.e. $n=3$

  $A\left(3\right)=1.2+\left(3-1\right)\left(-4\right)$

(Substitute 3 for n)

  $\begin{array}{l}A\left(3\right)=1.2+\left(2\right)\left(-4\right)\\ A\left(n\right)=1.2-8\\ A\left(n\right)=-6.8\end{array}$

(Simplify)

So, the third term is -6.8.

  $A\left(n\right)=1.2+\left(n-1\right)\left(-4\right)$

To find the seventh term, substitute the term number as 7 i.e. $n=7$

  $A\left(7\right)=1.2+\left(7-1\right)\left(-4\right)$

(Substitute 7 for n)

  $\begin{array}{l}A\left(7\right)=1.2+\left(6\right)\left(-4\right)\\ A\left(7\right)=1.2-24\\ A\left(7\right)=-22.8\end{array}$

(Simplify)

So, the seventh term is -22.8.

  $A\left(n\right)=1.2+\left(n-1\right)\left(-4\right)$

To find the tenth term, substitute the term number as 10 i.e. $n=10$ .

  $A\left(10\right)=1.2+\left(10-1\right)\left(-4\right)$

(Substitute 10 for n)

  $\begin{array}{l}A\left(10\right)=1.2+\left(9\right)\left(-4\right)\\ A\left(10\right)=1.2-36\\ A\left(10\right)=-34.8\end{array}$

(Simplify)

So, the tenth term is -34.8.