An auditorium has 40 rows. the first row has 20 seats and each row after has 2 additional seats. how many seats are there total?

This is an arithmetic sequence of the form:

a(n)=a+d(n-1), a=initial term, d=common difference, n=term number

The sum of an arithmetic sequence is the average of the first and last terms times the number of terms.

s(n)=(a+a+d(n-1))(n/2)

s(n)=(2a+dn-d)(n/2)

s(n)=(2an+dn^2-dn)/2

We are told that the initial term a=20 and that the common difference, d=2 so:

s(n)=(40n+2n^2-2n)/2

s(n)=(2n^2+38n)/2

s(n)=n^2+19n, so for 4o rows the total number of seats will be:

s(40)=40^2+19(40)

s(40)=1600+760

s(40)=2360

That’d be 9 years or something