Question

A certain theatre is shaped so that the first row has 23 seats and, moving towards the back, each successive row has 2 seats more than the previous row. If the last row contains 81 seats and the total number of seats in the theatre is 1560, how many rows are there?

Answer 1

There are
`30` rows in the theater.

Explanation:

Given terms.

Seats in order
`23,25,27....` will form an Arithmetic progression.

As the common difference
`d=2`,same for three consecutive terms so this an AP.

Now accordingly we can put the AP terms.

Total number of seats
`S_(n)=1560`

Seats in the first row
`a=23`

Seats in the last row
`l=81`

Summation of
`n` terms in AP
`(S_(n))`

`S_n={(n)/(2)[2a+(n-1)d]}`

OR

`S_n=(n)/(2)[a+l]`

Using the second equation.

`S_n=(n)/(2)[a+l]`

`1560=(n)/(2)[23+81]`

`n=(2* 1560)/(23+81)=(3120)/(104)=30`

So the number of rows (n) in the theater
`=30`