John makes a sequence where each stage is made up of a certain number of Xs arranged in a pattern - Junior Cycle Mathematics - Question 8 - 2016

To derive the formula, we observe the sequence:

Stage (S)	Number of Xs (N)
0	1
1	4
2	7
3	10

We notice that the number of Xs increases by 3 for each stage, leading us to deduce that:

$N = 4 + 3S$

Using the formula derived, we set it equal to 130:

$130 = 4 + 3k$

First, subtract 4 from both sides:

$126 = 3k$

Then, divide by 3:

$k = 42$

In Yoko’s sequence, using the formula $N = 1 + 2S$:

• Stage 0: $N = 1 + 2(0) = 1$ --> 1 X
• Stage 1: $N = 1 + 2(1) = 3$ --> 3 Xs
• Stage 2: $N = 1 + 2(2) = 5$ --> 5 Xs

The first three stages would look like:

Stage 0:  X
Stage 1:  X X X
Stage 2:  X X X X X

Using the established relationship:

$N = 1 + 2(y + 3)$

This simplifies to:

$N = 1 + 2y + 6 = 2y + 7$

Thus, the number of Xs in stage y + 3 of Yoko’s sequence is $2y + 7$.