SOLUTION: An audio-visual room (AVR) at Adamson University has 7 seats on the first row, 10 on the second row, 13 on the third row, and so on, and 40 rows of seats. How many seats are in the
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-> SOLUTION: An audio-visual room (AVR) at Adamson University has 7 seats on the first row, 10 on the second row, 13 on the third row, and so on, and 40 rows of seats. How many seats are in the
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Question 1197143: An audio-visual room (AVR) at Adamson University has 7 seats on the first row, 10 on the second row, 13 on the third row, and so on, and 40 rows of seats. How many seats are in the AVR? Answer by math_tutor2020(3817) (Show Source):
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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