Question 1197143
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Please see this similar question before reading onward
<a href = "https://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.1197145.html">https://www.algebra.com/algebra/homework/Sequences-and-series/Sequences-and-series.faq.question.1197145.html</a>
This similar question may help you solve the problem on your own. 
If you still need help, then the solution is in the next section below.


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We have an arithmetic sequence since the gap between terms is the same.
10-7 = 3
13-10 = 3
Each time we get to a new row, we add on 3 seats.
This is the common difference.
d = common difference = 3


The first term is a1 = 7


Summary so far
a1 = 7
d = 3


Let's form the nth term of the arithmetic sequence
an = a1 + d(n-1)
an = 7 + 3(n-1)


As a check, let's plug in n = 2 and we should get 10 seats
an = 7 + 3(n-1)
a2 = 7 + 3(2-1)
a2 = 10
That works out. I'll let you verify other values of n.


Now plug in n = 40
an = 7 + 3(n-1)
a40 = 7 + 3(40-1)
a40 = 124
There are 124 seats in row 40.


Now apply this summation formula
Sn = sum of the first n terms of an arithmetic sequence
Sn = (n/2)*(firstTerm + nthTerm) 
S40 = (40/2)*(firstTerm + 40thTerm) 
S40 = 20*(7 + 124)
S40 = 2620


An alternative formula is this
Sn = (n/2)*(2*a1 + d(n-1))
S40 = (40/2)*(2*7 + 3*(40-1))
S40 = 2620


The verification process is similar to that shown in the link above.


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Answer: <font color=red>2620</font> seats total
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