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?

Accepted Solution

Answer:There are [tex]30[/tex] rows in the theater.Step-by-step explanation:Given terms.Seats in order [tex]23,25,27....[/tex] will form an Arithmetic progression.As the common difference [tex]d=2[/tex],same for three consecutive terms so this an AP.Now accordingly we can put the AP terms.Total number of seats [tex]S_{n}=1560[/tex]Seats in the first row [tex]a=23[/tex]Seats in the last row [tex]l=81[/tex]Summation of [tex]n[/tex] terms in AP[tex](S_{n})[/tex][tex]S_n={\frac{n}{2}[2a+(n-1)d]}[/tex]OR[tex]S_n=\frac{n}{2}[a+l][/tex]Using the second equation.[tex]S_n=\frac{n}{2}[a+l][/tex][tex]1560=\frac{n}{2}[23+81][/tex][tex]n=\frac{2\times 1560}{23+81}=\frac{3120}{104}=30[/tex]So the number of rows (n) in the theater [tex]=30[/tex]