The number of seats per row in an auditorium increases from the front to the back. The first row has 15 seats, the second row has 2 more seats than the first row, the third row has 3 more seats than the second row, the fourth row has 2 more seats than the third row, the fifth row has 3 more seats than the fourth row. This pattern continues, with successive rows alternating between 2 more seats and then 3 more seats than the previous row. How many seats are in the auditorium if there are 30 rows total?

Answer:  The required number of seats in the auditorium is 1530.Step-by-step explanation:  Given that the number of seats per row in an auditorium increases from the front to the back. The first row has 15 seats, the second row has 2 more seats than the first row, the third row has 3 more seats than the second row, the fourth row has 2 more seats than the third row, the fifth row has 3 more seats than the fourth row. This pattern continues, with successive rows alternating between 2 more seats and then 3 more seats than the previous row. We are to find the number of seats in the auditorium if there are 30 rows in total. We know that, the sum of first n terms of an A.P. with first term a and common difference d is given by [tex]S_n=\dfrac{n}{2}\{2a+(n-1)d\}}.[/tex]According to the given information, we havethe number of seats in the 1st row = 15, the number of seats in the 2nd row = 15 + 2 = 17, the number of seats in the 3rd row = 17 + 3 = 20, the number of seats in the 4th row = 20 + 2 = 22, the number of seats in the 5th row = 22 + 3 =25, etc.So, the number of sets in the odd rows is an arithmetic progression with first term 15 and common difference 5. Also, total no. of odd rows is 15.That is, the number of seats in the odd rows is  [tex]n_o=\dfrac{15}{2}\{2\times15+(15-1)5\}=\dfrac{15}{2}(30+70)=750.[/tex]And, the number of sets in the even rows is an arithmetic progression with first term 17 and common difference 5. Also, total no. of even rows is 15.That is, the number of seats in the even rows is  [tex]n_e=\dfrac{15}{2}\{2\times17+(15-1)5\}=\dfrac{15}{2}(34+70)=780.[/tex]Therefore, the total number of rows in the auditorium is[tex]n=n_o+n_e=750+780=1530.[/tex]Thus, the required number of seats in the auditorium is 1530.