Answer:
1)
Number of Seats in next three rows: 37, 41, 45
Number of seats in row 21 = 105
2)
Cost for 4, 5, 6 miles:
Miles Cost($)
4 14.50
5 18.00
6 21.50
Cost for 12 miles = $42.50
Explanation:
1) The arithmetic sequence is
25, 29, 33 ....
Each term is 4 more than the previous term
So the next three terms starting with the 4th term area
33 + 4 = 37
37 + 4 = 41
41 + 4 = 45
In terms of the problem statement these are the number of seats in rows 4, 5, 6 respectively
The general equation for an arithmetic sequence nth term is
a(n) = a(1) + d(n - 1)
Here a(1) is the first term; here a(1) = 25
d = difference between successive terms called the common difference; here common difference = 4
n is of course the number of terms to be considered
using the values we have
a(n) = 25 + 4(n- 1)
= 25 + 4n - 4
= 21 + 4n
So the 21st term is a(21)
a(21) = 21 + 4 (21)
= 21 + 84
= 105
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2. Another arithmetic sequence
The first term is the charge for the first mile = 4
The second term = 7.5 for 2 miles
The third term = 11.5 for 3 miles
So d = 11 - 7.5 = 7.5 - 4 = 3.5
The cost for 4 miles = 11 + 3.50 = $14.50
The cost for 5 miles = 14.50 + 3.50 = $18
The cost for 6 miles = 18 + 3.50 = $21.50
Using the equation for finding the nth term we get
a(n) = a(1) + d(n - 1)
a(n) = 4 + 3.5(n-1)
a(n) = 4 + 3.5n - 3.5
a(n) = 0.5 + 3.5n
We have the following table
Miles Cost ($)
1 4.00
2 7.50
3 11.00
4 14.50
5 18.00
6 21.50
For 12 miles which would correspond to the 12 term
a(12) = 0.5 + 3.5(12)
= 0.5 + 42
= $42.50