Respuesta :
Answer:
Step-by-step explanation:
Given that Each row of the auditorium contains 2 more seats than the row in front of it. There are 25 seats at the comedian's level. We refer to the comedian-level row as Row 0, the row behind Row 0 as Row 1, etc. Let PN denote the number of seats in row N.
Thus no of seats in rows form an arithmetic progression with common difference =2
a = I term = P0 = 25
Hence general formula is [tex]P_n = 25+2n[/tex]
a) Find P1. Find P2.
27, 29
(b) The sequence P0, P1, P2, P3, ... is sequence.
Arithmetic
(c) How many seats are in Row 5?
[tex]25+2(5) = 35[/tex]
(d) How many seats are in Row 40?
[tex]25+40(2) = 105[/tex]
(e) Which is the first row with more than 90 seats?
[tex]25+2n>90\\2n>65\\n>32.5[/tex]
Row 33
(f) How many total seats are in Row 0 through Row 5?
[tex]25+27+...+35\\= \frac{6}{2} (25+35)\\= 180[/tex]
(g) There are 41 rows in the auditorium (labeled Row 0 through Row 40).
How many seats are in the entire auditorium?
Using Sum of 41 terms here
[tex]\frac{41}{2} (25*2+40*2) = 2665[/tex]
