Chapter 7.2, Problem 69PS

Solution

To identify: How many seats are in the front row of the theater? Write your answer in terms of n, x, and y.

A theater has n rows of seats, and each row has d more seats than the row in front of it. There are x seats in the last (nth) row and a total of y seats in the entire theater.

  $a_1=\frac{-nx+2y}{n}$ seats are in the front row of the theater.

Explanation:

There are $x$ seats in the last $nth$ row, hence

  $x=an=a_1+\left(n-1\right)d$

From (1) see that

  $a_1=x-\left(n-1\right)d$

Now observe the total number of seats in the entire theater is $y$ , so

  $y=s_n$

The general formula for sum $s_n=n\frac{\left(a_1+a_n\right)}{2}$

Put the value of $a_1=x-\left(n-1\right)d,a_n=x$ into $s_n=n\frac{\left(a_1+a_n\right)}{2}$

  $\begin{array}{l}y=n\frac{\left(x-\left(n-1\right)d+x\right)}{2}\\ 2y=n\left(2x-\left(n-1\right)d\right)\\ d=\frac{2nx-2y}{n\left(n-1\right)}\end{array}$

The front row is the first term of this sequence, so need to determine the term $a_1$

Put the value of $d$ into $a_1=x-\left(n-1\right)d$

  $\begin{array}{l}{a}_1=x-\left(n-1\right)d\\ a_1=x-\left(n-1\right)\frac{2nx-2y}{n\left(n-1\right)}\\ a_1=x-\frac{2nx-2y}{n}\\ a_1=\frac{xn-2nx+2y}{n}\\ a_1=\frac{-nx+2y}{n}\end{array}$