An arithmetic sequence has first term 2 and common difference 4. Another arithmetic sequence has first term 7 and common difference 5. Find the set of all numbers which are members of both sequences.

the mth term of the first sequence \( = 2 + 4(m - 1)\)     (M1)(A1)

the nth term of the second sequence \( = 7 + 5(n - 1)\)     (A1)

EITHER

equating these,     M1

\(5n = 4m - 4\)

\(5n = 4(m - 1)\)     (A1)

4 and 5 are coprime     (M1)

\( \Rightarrow 4|n\) so \(n = 4s\) or \(5|(m - 1)\) so \(m = 5s + 1\) , \(s \in {\mathbb{Z}^ + }\)     (A1)A1

thus the common terms are of the form \(\{ 2 + 20s;{\text{ }}s \in {\mathbb{Z}^ + }\} \)     A1

OR

the numbers of both sequences are

2, 6, 10, 14, 18, 22

7, 12, 17, 22     A1

so 22 is common     A1

identify the next common number as 42     (M1)A1

the general solution is \(\{ 2 + 20s;{\text{ }}s \in {\mathbb{Z}^ + }\} \)     (M1)A1

[9 marks]

Solutions to this question were extremely variable with some candidates taking several pages to give a correct solution and others taking several pages and getting nowhere. Some elegant solutions were seen including the fact that the members of the two sets can be represented as \(2\bmod 4\) and \(2\bmod 5\) respectively so that common members are \(2\bmod 20\).