An arithmetic sequence has a constant difference between consecutive terms. This is called the common difference $d$.
$n$th term formula: $$u_n = a + (n-1)d$$ where $a$ is the first term and $d$ is the common difference.
Example: Sequence 5, 8, 11, 14, … $a = 5$, $d = 3$ $$u_n = 5 + (n-1) \times 3 = 3n + 2$$
Check: $u_1 = 3(1) + 2 = 5$ ✓, $u_4 = 3(4) + 2 = 14$ ✓
Finding a term number: set $u_n$ equal to the target value and solve. $$3n + 2 = 200 \implies n = \frac{198}{3} = 66 \implies \text{200 is not in the sequence (not a whole number)}$$
Sum of arithmetic sequence: $$S_n = \frac{n}{2}(a + l) \quad \text{or} \quad S_n = \frac{n}{2}(2a + (n-1)d)$$ where $l$ is the last term.
Common error: writing the $n$th term as $a + nd$ (missing the $-1$), which gives a term shifted by one.
Solving linear equations Quadratic sequences
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2024 Jun 3H GCSE Q20 (2 marks) 2018 Jun 3H GCSE Q16 (1 mark)