Chapter 7.2, Problem 62MR

To determine

To Find: The third, seventh and tenth terms of the sequence described by the rule $A\left(n\right)=10+\left(n-1\right)\left(4\right)$

Answer to Problem 62MR

The required third, seventh and tenth term of the sequence is 18, 30 and 46 respectively.

Explanation of Solution

Given information:

The sequence $A\left(n\right)=10+\left(n-1\right)\left(4\right)$

Calculation:

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

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

  $A\left(3\right)=10+\left(3-1\right)\left(4\right)$

(Substitute 3 for n)

  $\begin{array}{l}A\left(3\right)=10+2\left(4\right)\\ A\left(3\right)=10+8\\ A\left(3\right)=18\end{array}$

(Simplify)

So, the third term is 18.

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

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

  $A\left(7\right)=10+\left(7-1\right)\left(4\right)$

(Substitute 7 for n)

  $\begin{array}{l}A\left(7\right)=10+5\left(4\right)\\ A\left(7\right)=10+20\\ A\left(7\right)=30\end{array}$

(Simplify)

So, the seventh term is 30.

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

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

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

(Substitute 10 for n)

  $\begin{array}{l}A\left(10\right)=10+9\left(4\right)\\ A\left(10\right)=10+36\\ A\left(10\right)=46\end{array}$

(Simplify)

So, the tenth term is 46.