Chapter 4.6, Problem 47E

To determine

whether the areas of the given figures form an arithmetic sequence or not and find the value of $f\left(30\right)$

Answer to Problem 47E

The value of $f\left(30\right)$ is $f\left(30\right)=117$

Explanation of Solution

Given:

  

Calculation:

The first term has one square

So, the area for that is

  $a_1=1{\text{ inch}}^2$

Similarly, for the second term have 5 square

So, the area for second term is

  $\begin{array}{l}{a}_2=5\left(1\right)\\ =5{\text{ inch}}^2\end{array}$

And third term has 9 square

So, the area for the third term is

  $\begin{array}{l}{a}_3=9\left(1\right)\\ =9{\text{ inch}}^2\end{array}$

Now, the common difference d is

  $\begin{array}{l}{a}_3-a_2=9-5\\ =4\end{array}$

And

  $\begin{array}{l}{a}_2-a_1=5-1\\ =4\end{array}$

Therefore, the sequence is arithmetic.

The first term is 1, and the common difference is 4.

Consider the function for the arithmetic sequence

  $f\left(n\right)=a_1+\left(n-1\right)d$

Put the value $a_1=1$ and $d=4$

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

Therefore, the function that represents the arithmetic sequence is $f\left(n\right)=4n-3$ .

Now, the 30th term is

  $\begin{array}{l}f\left(30\right)=4\left(30\right)-3\\ =120-3\\ =117\end{array}$

Conclusion:

Therefore, the required value is $f\left(30\right)=117$ .