The prompt was devised by Helen Hindle, a Lead Practitioner in maths. It generalises from a student's observation made during the intersecting sequences inquiry (see box right). The student proved the statement is true for the sequence generated from 6n + 1. Teachers could use the prompt as the start of a standalone inquiry or introduce it as a separate pathway during the intersecting sequences inquiry. The questions and observations that develop from the statement include:  Is it true or false?
 If it is true, then is it true for some or all sequences?
 3 x 5 = 15 and 3, 5 and 15 are in the sequence generated by 2n + 1.
 Does it only work with consecutive terms?
 Are there any sequences when you could use the sum of the terms?
 How could you show this is always true?
 What would happen if you multiplied terms from two sequences? Would the answer be in both sequences?
Students realise that most expressions do not generate arithmetic sequences for which the prompt is true. During exploration, they conclude that the prompt is true for any expression ending with +1. It is also true for any expression in the form an + a, such as 2n + 2, 3n + 3 and so on.
Counterexample and proof Counterexample to the prompt Using 3n  1 as the expression for the n^{th} term, we get 2, 5, 8, 11, 14. In a general form, we have:3k  1, 3(k + 1)  1, 3(k + 2)  1, 3(k + 3)  1. The product of, say, the second and fourth term is [3(k + 1)  1][3(k + 3)  1] = 9k^{2 }+ 30k + 16. This can be written as 3(3k^{2} + 10k) + 16, which is not in the form 3n  1. Proof The prompt is true for any expression of the n^{th} term that has a constant of one (for example, 6n + 1). The general sequence starts: ak + 1, a(k + 1) + 1, a(k + 2) + 1, a(k + 3) + 1. Whichever two terms we choose, the expression for the product of the two will always end with +1, which corresponds to the general form of an + 1. Proof If the coefficient and constant in the expression are the same, then the prompt is true. The general sequence starts: ak + a, a(k + 1) + a, a(k + 2) + a, a(k + 3) + a. Whichever two terms we choose, the expression for the product of the two will always end with a term in a^{2}. This can then be rearranged to give a as the final term. The product of the first two terms, for example, is a[(ak^{2} + 3ak + a) + a].
 Conjecture and proof George Marsden (a year 10 student at St. Andrew's School, Leatherhead, UK) proved his conjecture that the product of any two terms in the sequence given by the expression for the nth term 6n + 1 is also a term in the same sequence. For example, the product of 7 and 13 (both terms in the sequence generated from 6n + 1) is 91, which is the fifteenth term in the sequence. The illustration below shows how George proved his conjecture.
