Chapter 9.4, Problem 46PPSE

Solution

a.

To write an arithmetic series for the number of seats in the theater.

  $18+21+24+\mathrm{...}+75$

Given information:

There are 20 rows in the meeting room and the first row has 18 seats, and the number of seats increases by 3 with each successive row.

Calculation:

The sequence is an arithmetic sequence with $a_1=12$ and $d=3$ .

Find the nth term of the sequence.

  $\begin{array}{c}{a}_n=a_1+\left(n-1\right)d\\ =18+\left(n-1\right)3\\ =3n+15\end{array}$

Therefore, the number of seats in each row can be found using the explicit formula, $a_n=3n+15$ .

Find the number of seats in the 20^th row.

  $\begin{array}{c}{a}_{20}=3\cdot 20+15\\ =60+15\\ =75\end{array}$

Therefore, there are 75 seats in the 20^th row.

Thus, the series can be written as $18+21+24+\mathrm{...}+75$ .

b.

To find the total seating capacity of the theater.

The total seating capacity of the theater is 930.

Given information:

There are 20 rows in the meeting room and the first row has 18 seats, and the number of seats increases by 3 with each successive row.

Calculation:

The first term of the series is 18 and the 20^th term is 75.

Find the sum of the series using the formula, $S_n=\frac{n}{2}\left(a_1+a_n\right)$ .

  $\begin{array}{c}{S}_{20}=\frac{20}{2}\left(a_1+a_{20}\right)\\ =\frac{20}{2}\left(18+75\right)\\ =10\left(93\right)\\ =930\end{array}$

Therefore, the total seating capacity of the theater is 930.

c.

To find the amount of money generated by a full-house.

The total amount generated is $46950.

Given information:

There are 20 rows in the meeting room and the first row has 18 seats, and the number of seats increases by 3 with each successive row.

Calculation:

Find the total number of seats in the 5^th, 10^th and 15^th rows.

  $\begin{array}{c}{a}_5=3\cdot 5+15\\ =15+15\\ =30\\ a_{10}=3\cdot 10+15\\ =30+15\\ =45\\ a_{15}=3\cdot 15+15\\ =45+15\\ =60\end{array}$

Now, find the sum of series till 5^th row, 10^th row and 15^th row.

  $\begin{array}{c}{S}_5=\frac{5}{2}\left(a_1+a_5\right)\\ =\frac{5}{2}\left(18+30\right)\\ =5\cdot 24\\ =120\\ S_{10}=\frac{10}{2}\left(a_1+a_{10}\right)\\ =\frac{10}{2}\left(18+45\right)\\ =5\cdot 63\\ =315\\ S_{15}=\frac{15}{2}\left(a_1+a_{15}\right)\\ =\frac{15}{2}\left(18+60\right)\\ =15\cdot 39\\ =585\end{array}$

Now, the find the amount generated as follows:

  $\begin{array}{c}\text{Amount}=60{\displaystyle \sum _{n=1}^5\left(3n+15\right)}+55{\displaystyle \sum _{n=6}^{10}\left(3n+15\right)}+50{\displaystyle \sum _{n=11}^{15}\left(3n+15\right)}+45{\displaystyle \sum _{n=16}^{20}\left(3n+15\right)}\\ =60\cdot S_5+55\left(S_{10}-S_5\right)+50\left(S_{15}-S_{10}\right)+45\left(S_{20}-S_{15}\right)\\ =60\cdot 120+55\left(315-120\right)+50\left(585-315\right)+45\left(930-585\right)\\ =\text{46950}\end{array}$

Therefore, the total amount generated is $46950.