Respuesta :
Let n be the total number of rows and there are x seats is the 1st row where x is an odd natural number.
As each row must have 2 more seats than the one before it, so
Number of seats in 2nd row = x+2
Number of seats in 3rd row = x+2+2=x+2x2
Number of seats in 4th row = x+2+2+=x+2x3
Similarly, the number of seats in (n-1)th row =x+2x(n-1)
the number of seats in n^{th} row, i.e last row  =x+2x n.
As the front row must have one-quarter of the total seats in the back 2 rows combined, so
[tex]x =\frac 1 4 ( x+2\times(n-1) + x+2\times n) \\\\\Rightarrow 4x=2x+4n-6 \\\\\Rightarrow 4x-2x=4n-6 \\\\\Rightarrow 2x=4n-6 \\\\\Rightarrow x=2n-3\cdots(i)[/tex]
So, the number of seats in the 1st row, x, and the total number of seats, n, must satisfy the equation (i).
For [tex]x>0, n\geq 3.[/tex]
So, for n=3 rows
The number of seats in the 1st row, x= 2x3-3=3.
The number of seats in the 2nd row, = 3+2=5 as in subsequent rows, there will be 2 more seats.
and the number of seats in the 3rd row (last row)=5+2=7.
n can have any integral values satisfying  [tex]n\geq 3[/tex].
