Step-by-step explanation:
a) We can find the iterative rule by first making a table of values.
n -> [tex]a_n[/tex]
1 -> 22
2 -> 28
3 -> 34
We see that [tex]a_n[/tex] increases by 6 each time. Hence, the iterative rule should have the term 6n. However, if we put 1 in for n, we get 6. We want 22-6=16 more than this, or [tex]6n+16[/tex]. This would work for any seat row we give.
b) Our iterative rule to get the number of seats is [tex]6n+16[/tex]. Since we know that this value is 100, we can put both equal to each other.
[tex]6n+16=100\\6n=84\\n=14[/tex]
Thus, the 14th row has 100 seats.
Answer:
(a) S_n = S_1 + 6 (n - 1)
(b) Row 14 has 100 seats
Step-by-step explanation:
(a) The arithmetic sequence would follow the iterative rule,
S_n = S_1 + 6 (n - 1) where S is the number of seats in row n, n is the row number, and S is the number of seats in row 1.
Row 1 = S_1 = 22
2 = S_2 = 28
3 = S_3 = 34
4 = S_4 = 40
|| || = 46
|| || = 52
|| || = 58
|| || = 64
|| || = 70
|| || = 76
|| || = 82
|| || = 88
|| || = 94
Row 14 = S_14 = 100
(b) 100=22+6(n-1) solve for n after setting S_n = 100
100 = 22 + 6_n - 6
84/6 = 6n/6 Row 14 has 100 seats
14 = n
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