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Junior/Intermediate Sequences Tips Tip 1: It may help to use algebra Question: The first two terms of a sequence are positive integers. Each subsequent term is the sum of the previous two terms. The 6th term of the sequence is 50. What are the possible values of the first term in the sequence? Since the first two terms are arbitrary, we could represent them as variables 𝑎 and 𝑏. Then using the definition of the sequence, our terms are 𝑎, 𝑏, 𝑎 + 𝑏, 𝑎 + 2𝑏, 2𝑎 + 3𝑏, 3𝑎 + 5𝑏. Thus 3𝑎 + 5𝑏 = 50. This is an example of a Linear Diophantine Equation (see Number Theory tips). A ‘Diophantine Equation’ is an equation where we’re interested in integer solutions. It’s ‘linear’ because it looks like the equation of a straight line (just as if we had something like 3𝑦 = 𝑥 + 2). Let’s try a few values of 𝑏. o When 𝑏 = 1, we find that 𝑎 = 15. This gives us one solution. o When 𝑏 = 2, then we have 3𝑎 = 40, so 𝑎 cannot be an integer. o Continuing on, we also get 𝑏 = 4 and 𝑏 = 7. o There can’t be any further solutions, because when 𝑏 = 10, then 𝑎 = 0 which is not positive. To avoid losing marks in an Olympiad, you should justify why there are no further solutions than the ones you have found. Tip 2: Spot when your sequence cycles. Suppose in following instructions to generate a sequence, you obtain 9, 3, 6, 9, 3, 6, 9, 3, 6, … What would be the 1000th term? In such cases, we just need to be methodical. We can see that any term where the position is a multiple of 3 will be 6 (i.e. the 3rd term, 6th term, etc.). Thus the 999th term will be 6, and thus the 1000th term will be 9. www.drfrostmaths.com