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Please help with this!

Please help with this!-example-1
User Ziv Galili
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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

------------------------------------------

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

User TheRealEmu
by
8.7k points

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