Chapter 8.2, Problem 76E

To determine

To find: the number of seats in all 20 rows.

Answer to Problem 76E

There are 590 seats in 20 rows.

Explanation of Solution

Given information:

An auditorium has 20 rows of seats. Therefore are 20 seats in the first row, 21 seats in the second row, 22 seats in the third row.

Given figure,

  

Concept used:

An arithmetic sequence of n terms, has the form $a_1,a_2,a_3,\mathrm{...},a_n$ and should have a common difference between each two consecutive terms.

That is

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}$

Common difference can be defined by d .

  $a_2-a_1=a_3-a_2=a_4-a_3=\mathrm{...}=a_n-a_{n-1}=d$

nth term of the arithmetic sequence has the form

  $a_n=a_1+\left(n-1\right)d$

Where $a_1$ is the first term of the sequence, and d is the common difference.

Sum of an arithmetic finite sequence has the form

  $S_n=\frac{n}{2}\left(a_1+a_n\right)$

Here, n is number of terms, $a_1$ is the first term of the sequence, and $a_n$ is the last term of the sequence.

Calculation:

An auditorium has 20 rows of seats there are 20 seats in a first row, 21 seats in the second row.

  $\begin{array}{l}{a}_1=20\\ a_2=21\end{array}$

The common difference between these seats is

  $\begin{array}{c}d=a_2-a_1\\ =21-20\\ =1\end{array}$

Therefore are 20 rows of seats using formula $S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$

  $\begin{array}{c}{S}_{20}=\frac{20}{2}\left(2\times 20+\left(20-1\right)1\right)\\ =10\left(40+19\right)\\ =10\times 59\\ =590\end{array}$

Hence, there are 590 seats in 20 rows respectively.