Sile is investigating the number of grey square tiles needed to make patterns in a sequence - Leaving Cert Mathematics - Question 7 - 2014

The sequence of the number of tiles is an arithmetic sequence with a first term of 21 and a common difference of 12. Thus, the formula for the nth pattern can be expressed as:

$T_n = 21 + (n-1) imes 12 = 12n + 9$

Substituting n = 10 into the formula:

$T_{10} = 12(10) + 9 = 120 + 9 = 129$

Thus, the tenth pattern requires 129 tiles.

We need to find the largest whole number n such that:

$T_n \leq 399$

This leads to the inequality:

$12n + 9 \leq 399\quad \Rightarrow \quad 12n \leq 390 \quad \Rightarrow \quad n \leq 32.5$

Therefore, the largest integer n is 32, meaning the biggest pattern that Sile can make is pattern 32.

To find the total number of tiles used in the first n patterns, we can use the formula for the sum of an arithmetic series:

$S_n = \frac{n}{2} [2a + (n-1)d]$ where a = 21 and d = 12. Thus, the formula for the total number of tiles in the first n patterns is:

$S_n = \frac{n}{2} [2(21) + (n-1)(12)] = 6n^2 + 15n$

We want to find the largest n for which:

$S_n \leq 399$ Substituting our formula, we set up the inequality:

$6n^2 + 15n \leq 399$ This simplifies to:

$2n^2 + 5n - 133 \leq 0$ We can factor this quadratic equation to find n. The potential candidates are:

• The positive roots of the function show that n = 7 is the maximum since S_7 < 399 (as S_7 = 210) and S_8 exceeds it. Thus, Sile can make 7 patterns in total.