A patio was to be laid in a design with one tile in the
first row, two tiles in the second row, three
tiles in the third row, and so on. Mr. Tong had 60
tiles to use. How many tiles should be placed
in the bottom row to use the most tiles?

We can model this as an arithmetic series, where the first term is 1 and the common difference is 1. The sum of the first n terms of an arithmetic series is:

S = n/2 (2a₁ + d (n − 1))

Given that a₁ = 1 and d = 1:

S = n/2 (2(1) + n − 1)

S = n/2 (n + 1)

Since S ≤ 60:

n/2 (n + 1) ≤ 60

n (n + 1) ≤ 120

n must be an integer, so from trial and error:

n ≤ 10

Mr. Tong should use 10 tiles in the final row to use the most tiles possible.

A patio was to be laid in a design with one tile in the first row, two tiles in the second row, three tiles in the third row, and so on. Mr. Tong had 60 tiles to use. How many tiles should be placed in the bottom row to use the most tiles?

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A patio was to be laid in a design with one tile in the
first row, two tiles in the second row, three
tiles in the third row, and so on. Mr. Tong had 60
tiles to use. How many tiles should be placed
in the bottom row to use the most tiles?