Question

A theater has 100 seats in the first row, 120 seats in the second row, 140 seats in the third row, and so on in the same increasing pattern. If the theater has 40 rows of seats, how many seats are in the theater?

• A. $176000$
• B. $18600$
• C. $19600$
• D. $20600$

Answer 1

Answer: (C)

$19600$

The seating pattern is forming an arithmetic sequence: $100,120,140,..$
First term: $a=100$
$S_n=?$
$d=20$
$n=40$
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$S_{40}=\frac{40}{2}[2\times 100+(40-1\left)20\right]$

$S_{40}=20[200+780]$

$S_{40}=19600$

There are $19600$ seats.