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2 Responses
You’re on the right track noticing the gaps 4, 6, 8, 10 increase by 2; that means the sequence has a constant second difference, so it should come from a quadratic rule a_n = an^2 + bn + c. For quadratic sequences, the constant second difference equals 2a, so 2a = 2 gives a = 1. Now use the first terms to solve for b and c: with n = 1, 1 + b + c = 3 so b + c = 2; with n = 2, 4 + 2b + c = 7 so 2b + c = 3; subtracting gives b = 1 and then c = 1. Therefore the nth term is a_n = n^2 + n + 1. Your “next gap is 12” fits this: the gaps follow 2n + 2 (so a recursive rule is a_{n+1} = a_n + 2n + 2), and at n = 5 that gives 12, making the next term 31 + 12 = 43. For more on spotting quadratic sequences via second differences, this overview is helpful: https://www.khanacademy.org/math/algebra/sequences/alg2-second-differences Hope this helps!
Spot the drumbeat: the gaps 4, 6, 8, 10 are consecutive even numbers (each step grows by 2), so the sequence is quadratic with a_n = n^2 + n + 1 (n=1,2,3,…) and the next term is 31 + 12 = 43. For a quick tour of “second differences” and quadratic sequences, see https://www.bbc.co.uk/bitesize/guides/zp8mrdm/revision/1 – hope this helps!