June 2024 Paper 2 Q4
4 Here is some information about three linear sequences, \(A\), \(B\) and \(C\).
The \(n\)th term of \(C\) is \(42 - 3n\)
The first four terms of \(B\) are
14 22 30 38
\[n\text{th term of } C = n\text{th term of } A + n\text{th term of } B\]Work out the 20th term of \(A\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: works out \(n\)th term of B | ||
| Common difference of 8 or \(8n\) | M1 | implied by \(8(n - 1)\) |
| \(8n + 6\) | M1dep | oe eg \(14 + 8(n - 1)\) |
| \(42 - 3n -\) their \((8n + 6)\) or \(36 - 11n\) | M1 | oe expression for \(n\)th term of A their \(8n + 6\) must be of the form \(an + b\) with \(a\) and \(b\) non-zero missing brackets must be recovered \(36 - 11n\) or \(36 - 11 \times 20\) is M3 |
| \(-184\) | A1 | |
| Alternative method 2: works out 20th term of B | ||
| Common difference of 8 or \(8n\) | M1 | implied by \(8(n - 1)\) |
| \(14 + 8 \times 19\) or 166 | M1dep | oe eg \(38 + 8 \times 16\) |
| \(42 - 3 \times 20\) or \(-18\) | M1 | |
| \(-184\) | A1 | |
| Alternative method 3: works out consecutive terms of C | ||
| 39 36 | M1 | |
| \(39 - 14\) and \(36 - 22\) or 25 and 14 | M1dep | oe |
| \(25 - 11 \times 19\) | M1dep | oe |
| \(-184\) | A1 | |
Additional guidance
| Up to M3 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Condone \(-184\) seen in working with answer 184 | M3A1 |
| 1st M1 Common difference of 8 may be seen by the sequence for B | |
| Alt 3 may be seen using any pair of consecutive terms of C |