Write the first five terms of the sequence \(3n + 4\).

\(n\) represents the position in the sequence. The first term in the sequence is when \(n = 1\), the second term in the sequence is when \(n = 2\), and so on.

To find the terms, substitute \(n\) for the position number:

• when \(n = 1\), \(3n + 4 = 3 \times 1 + 4 = 3 + 4 = 7\)
• when \(n = 2\), \(3n + 4 = 3 \times 2 + 4 = 6 + 4 = 10\)
• when \(n = 3\), \(3n + 4 = 3 \times 3 + 4 = 9 + 4 = 13\)
• when \(n = 4\), \(3n + 4 = 3 \times 4 + 4 = 12 + 4 = 16\)
• when \(n = 5\), \(3n + 4 = 3 \times 5 + 4 = 15 + 4 = 19\)

The first five terms of the sequence: \(3n + 4\) are \(7, 10, 13, 16, 19, ...\)

The nth term for a quadratic sequence has a term that contains \(x^2\). Terms of a quadratic sequence can be worked out in the same way.

Write the first five terms of the sequence \(n^2 + 3n - 5\).

• when \(n = 1\), \(n^2 + 3n - 5 = 1^2 + 3 \times 1 - 5 = 1 + 3 – 5 = -1\)
• when \(n = 2\), \(n^2 + 3n - 5 = 2^2 + 3 \times 2 - 5 = 4 + 6 – 5 = 5\)
• when \(n = 3\), \(n^2 + 3n - 5 = 3^2 + 3 \times 3 - 5 = 9 + 9 – 5 = 13\)
• when \(n = 4\), \(n^2 + 3n - 5 = 4^2 + 3 \times 4 - 5 = 16 + 12 – 5 = 23\)
• when \(n = 5\), \(n^2 + 3n - 5 = 5^2 + 3 \times 5 - 5 = 25 + 15 – 5 = 35\)

The first five terms of the sequence: \(n^2 + 3n - 5\) are -1, 5, 13, 23, 35