October 2021 Paper 2 Q3
3 The 15th term of an arithmetic sequence is 88. The sum of the first 10 terms is 310.
Determine the first term and the common difference. [6]
| Scheme | Marks |
|---|---|
| \(a + 14d = 88\) (i) | M1 A1 |
| \(\frac{10}{2}(2a + 9d) = 310\) (ii) | M1 A1 |
| Substitute from (i) into (ii) | M1 |
| \(\frac{10}{2}(2(88 - 14d) + 9d) = 310\) \(176 - 19d = 62\) \(d = 6\), \(a = 4\) | A1 |
| [6] |
Notes
M1 for one error, eg \(a + 9d = 88\)
M1 for \(\frac{15}{2}(2a + 14d) = 310\) or for one error, eg \(5(2a + 18d) = 310\)
M1: Attempt substitution, elimination or to verify solutions found BC.
Condone one arithmetical error
SC for last 2 marks: Correct answers, substitution not seen: B1B0, dep M1M1
A1: cao