Gertie writes down the following sequence, which repeats every three terms: 3, 6, 4, 3, 6, 4, 3, .. - Junior Cycle Mathematics - Question 9 - 2019

To find the 100th term, calculate:

100 mod 3 = 1

Since the result is 1, the 100th term is the same as the 1st term, which is 3.

To find the n-th term, determine:

• If n modulo 3 equals 1, the term is 3 (1st term).
• If n modulo 3 equals 2, the term is 6 (2nd term).
• If n modulo 3 equals 0, the term is 4 (3rd term).

For the sequence starting with 8:

1. 2nd term: $1/2 (8 + 2) = 5$
2. 3rd term: $2 imes 5 = 10$
3. 4th term: $1/2 (10 + 2) = 6$
4. 5th term: $2 imes 6 = 12$

Thus, the terms are 8, 5, 10, 6, and 12.

Ahmed's sequence will stay constant at 2. This is because every term derived from 2 will always produce the next term as 2, thus resulting in a sequence of 2, 2, 2, ... indefinitely.

Considering 86 as the 2nd term:

1. If the 1st term is even, the calculation is: rac{1}{2}(2nd erm - 2) = rac{1}{2}(86 - 2) = 42 The 1st term can be 42.
2. If the 1st term is odd, the calculation is:

Given the first term is k (which is odd):

1. 2nd term: $2k$
2. 3rd term: $1/2(2k + 2) = k + 1$
3. 4th term: $2(k + 1) = k + 3$

Thus, the next three terms are $2k$, $k + 1$, and $k + 3$.