Question

In the audiotrium, the number of seats per row increases at a constant rate. Therer are 28 seats in row 2 and 37 seats in row 5. How many seats are there in row 15?

Answer 1

67 seats

Explanation:

We are going to use the arithmetic progression to solve this problem

remember that the the formula is

a + (n -1)d = n

Where n is the nth term

A is the first term

D is the difference

Apply the given variables into this equation and we have

In row 2 ,we have a total of 28 seats

And in row 5, we have 37 seats

This will lead to

a +(n - 1)d = 37 and here n is the 5th term so we have

a +4d = 37____ first equation

For row 2,

a + (n - 1)d = 28(remember that the nth term here is 2)

And we have

a + (2 - 1)d = 28 which is:

a + d = 28_____ equation 2

Now bring the 2 equations together and we have

a+4d= 37____ equation 1

a+d = 28_____ equation 2

Subtract the 2nd term from the 5th term. This will be:

3d= 9

d= 3

Common difference is 3

Since d= 3, a will be

a+d = 28

a+3= 28

a= 28-3 = 25

Therefore,the number of seats in row 15 will be

a+14d

= 25+(14×3)

= 25 + 42

= 67 seats in row 15

Answer 2

Answer: 67 seats

Explanation:

Number of seats in row 2= 28

Number if seats in row 5 =37

This is a form of arithmetic progression.

Subtract the 2nd term from the 5th term. This will be:

a+4d= 37

-a+d = 28

3d= 9

d= 9/3

d= 3

Common difference is 3

Since d= 3, a will be

a+d = 28

a+3= 28

a= 28-3 = 25

Number of seats in row 15 will be:

a+14d

= 25+(14×3)

= 25 + 42

= 67

Row 15 has 67 seats