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

Answer:

There are $30$ rows in the theater.

Step-by-step 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$(S_{n})$

$S_n={\frac{n}{2}[2a+(n-1)d]}$

OR

$S_n=\frac{n}{2}[a+l]$

Using the second equation.

$S_n=\frac{n}{2}[a+l]$

$1560=\frac{n}{2}[23+81]$

$n=\frac{2\times 1560}{23+81}=\frac{3120}{104}=30$

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