Cambridge IGCSE Maths 0580 Sequences Questions

1. The terms go up by \(6\) each time.
2. \(26 + 6 = 32\), \(32 + 6 = 38\)

Answer: (a) \(32, 38\) (b) add \(6\)

1. The difference is \(3\), so the \(n\)th term starts \(3n\).
2. \(3 \times 1 = 3\) and the first term is \(5\), so add \(2\).

Answer: \(3n + 2\)

• B1 (a) 4, 7, 12
• B1 (b) 103

Answer: (a) \(4, 7, 12\) (b) \(103\)

1. \(4n - 1 = 99\) gives \(4n = 100\), \(n = 25\).
2. \(25\) is a whole number, so \(99\) is the \(25\)th term.

Answer: Yes, \(99\) is the \(25\)th term

1. Compare with the square numbers \(1, 4, 9, 16, 25\): each term is \(2\) more.

Answer: \(n^2 + 2\)

• M1 compares with n²: 1, 4, 9, 16, 25, or second difference 2
• A1 n² − 1 oe

Answer: \(n^2 - 1\)

1. First differences: \(7, 11, 15\). Second difference: \(4\), so the sequence starts \(2n^2\).
2. Subtract \(2n^2\) (\(2, 8, 18, 32\)): \(3, 4, 5, 6\), which is \(n + 2\).

Answer: \(2n^2 + n + 2\)

1. (a) Each term is double the one before, starting at \(3\): \(3 \times 2^{n-1}\).
2. (b) \(3 \times 2^7 = 3 \times 128 = 384\)

Answer: (a) \(3 \times 2^{n-1}\) (b) \(384\)

1. (a) Compare with the cube numbers \(1, 8, 27, 64\): each term is \(1\) more, so \(T_n = n^3 + 1\).
2. (b) \(T_{10} = 1000 + 1 = 1001\)

Answer: (a) \(T_n = n^3 + 1\) (b) \(1001\)

• M1 3n² − 2n = 645
• M1 solves 3n² − 2n − 645 = 0, e.g. (3n + 43)(n − 15) = 0 or formula
• A1 15th term

Answer: the 15th term