Question

In: Math

A community baseball stadium has 10 seats in the first row, 13 seats in the second row, 16 seats in the third row, and ...

A community baseball stadium has 10 seats in the first row, 13 seats in the second row, 16 seats in the third row, and so on. There are 56 rows in all. What is the seating capacity of the stadium?

Expert Solution

A community baseball stadium has 10 seats in the first row, 13 seats in the second row, and 16 seats in the third row and so on.

The number of seats in different rows can be written in a sequence as follows:

10, 13, 16, …

As each term of the sequence is obtained by adding 3 to the previous term, the above sequence is an arithmetic sequence with common difference d = 3

First term of the sequence is a1 = 10

There are 56 rows in all. Hence, n = 56

Use the formula for nth term of an arithmetic sequence,

an = a1 + (n – 1)d ...... (1)

Substitute a1 = 10, d = 3 and n = 56 in (1) and compute 56th term of the sequence,

a56 = 10 + (56 – 1) × 3

= 10 + 55 × 3

= 10 + 165

= 175

Use the formula for sum of first n terms of an arithmetic sequence,

Sn = n(a1 + an)/2 ...... (2)

Substitute a1 = 10, a56 = 175 and n = 56 in (2) and simplify,

S56 = 56(10 + 175)/2

= 56 × 185/2

= 5180

Therefore, the seating capacity of the stadium is 5180.

Junaid answered 1 year ago

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