Answer:
(a)14 seats
(b)1,050 seats
Step-by-step explanation:
The theater has 10 seats in the first row and 30 seats in the 6th row.
We can model this as an arithmetic progression problem where:
• The first term, a = 10
,
• The last term, l = 30 when n=6
We know that for an arithmetic progression:
![\begin{gathered} l=a+(n-1)d \\ 30=10+(6-1)d \\ 30=10+5d \\ 5d=30-10 \\ 5d=20 \\ d=(20)/(5)=4 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/otgv8scu3e3muhgulukxxe5es3f8fxvnth.png)
Therefore, the number of seats in the 11th row will be:
![a_(11)=10+4=14\text{ seats}](https://img.qammunity.org/2023/formulas/mathematics/college/u1r91k3xmwucfnyzbogl6zb8mdoopvspcf.png)
(b)The theater has a total of 21 rows.
To determine the total number of seats, we use the formula for the sum of an arithmetic progression.
![S_n=(n)/(2)(2a+(n-1)d)](https://img.qammunity.org/2023/formulas/mathematics/college/999tr0bbwd6nzd4ivmm8898rpj22q6e4cj.png)
We define the variables:
• Since the theatre has a total of 21 rows, therefore n=21
,
• The first term, a = 10
,
• Common difference, d = 4
We substitute into the formula above:
![\begin{gathered} S_n=(n)/(2)(2a+(n-1)d)\implies S_(21)=(21)/(2)(2*10+(21-1)*4) \\ =10.5(20+20*4) \\ =10.5(20+80) \\ =10.5*100 \\ =1050 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/c0wz63c834rg7ckt5756nwirbrlma20iye.png)
There are a total of 1,050 seats in the theater.