Question

A theater has 30 seats in the first row. Each row behind the first row

Answer 1

Answer:

a, Row 1 = 30 seats

Row 2 = 32 seats

Row 3 = 34 seats

Row 4 = 36 seats

b, n + equidistant number of seats added each row (2) * number of row - 1

c, 58 seats

Step-by-step explanation:

a, Because the seats is increasing by 2 after every row after row 1

Therefore, row 2 = row 1 + 2 = 30+2 = 32 seats

Row 3 = Row 2 + 2 =32+2 =34 seats

Row 4 = Row 3 + 2 = 34 + 2 = 36 seats

b, The explicit rule for the number of seats would be:

Row 1 = n

Row 2 = n + equidistant number of seats added each row (2) * (2-1) = n + 2

Row 3 = n+ equidistant number of seats added each row (2) * (3-1) = n+ 4

Row 4 = n + equidistant number of seats added each row(2) * (4-1) = n+6

etc.....

c, The number of seats in the 15th row would be = number of seat in the first row + equidistant number of seats added each row * 15 - 1 (first row)

= 30 + 2 * 14= 30 + 28 = 58 seats

Hope this help you :3

Answer 2

Answer:

a). 30, 32, 34, 36.....

b). $A_{n}=A_{0}+(n-1)d$

c). There are 58 seats in the the 15th row.

Step-by-step explanation:

A theater has 30 seats in the first row.

Each row behind the first row gains two additional seats.

a). Each row of theater will be

30, 32, 34, 36........

b). Since sequence formed is an arithmetic sequence so the explicit formula of the sequence will be

$A_{n}=A_{0}+(n-1)d$

where $A_{n}$ = nth term of the sequence

$A_{0}$ = first term of the sequence

n = number of term

d = common difference

Now w substitute the values in the explicit formula

$A_{n}$=30 + (n - 1)2

$A_{n}$ = 30 + 2n - 2

$A_{n}$ = 2n + 28

c). Now we have to find the value of $A_{15}$

$A_{15}$ = 2×15 + 28 = 58

So there are 58 seats in the 15th row of the theater.