Chapter 12.1, Problem 56E

To determine

To calculate: Thedistance travelled by her if she reached the finished line in 15 seconds.

Answer to Problem 56E

The distance travelled by her is $810\text{ feet}$ .

Explanation of Solution

Given information:

The distance travelled in first second is 5 feet and for each subsequent second the distance increased by 7 feet.

Formula used:

A sequence with first term as $a_1$ and next term written as sum of preceding term and common difference then it is called an arithmetic sequence. It is denoted as $a_1,a_1+d,a_1+2d+\dots$ .

The sum of n terms of the arithmetic sequence is $S_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]$ .

Calculation:

Consider the provided information that distance travelled in first second is 5 feet and for each subsequent second the distance increased by 7 feet.

The above situation forms an arithmetic sequence as a constant increase in distance is there for each subsequent second.

Denote the distance travelled by her if she reached the finished line in 15 secondsby $S_{15}$ .

Recall that a sequence with first term as $a_1$ and next term written as sum of preceding term and common difference then it is called an arithmetic sequence. It is denoted as $a_1,a_1+d,a_1+2d+\dots$ .

The sum of n terms of the arithmetic sequence is $S_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]$ or $S_n=\frac{n}{2}\left[a_1+a_n\right]$

Here $a_1=5,d=7\text{ and }n=15$ .

Again apply the sum of n terms of the arithmetic sequence $S_n=\frac{n}{2}\left[2a_1+\left(n-1\right)d\right]$ ,

  $\begin{array}{c}{S}_{15}=\frac{15}{2}\left[2\left(5\right)+\left(15-1\right)7\right]\\ =\frac{15}{2}\left[10+98\right]\\ =\frac{15}{2}\left(108\right)\\ =810\end{array}$

Thus, the distance travelled by her if she reached the finished line in 15 secondsis $810\text{ feet}$ .