Foundation June 2025 Paper 3 Q12
12 The \(n\)th term of a sequence is \(4n - 1\)
(a) Work out the 3rd term of the sequence. (1)
Here are the first four terms of a different sequence.
9 15 21 27
(b) Is 63 a number in this sequence?
You must give a reason for your answer. (2)
You must give a reason for your answer. (2)
| Answer | Mark | Mark scheme |
|---|---|---|
| 11 | B1 | cao |
| Answer | Mark | Mark scheme |
|---|---|---|
| Yes (supported) | M1 | for method to show that 63 is in the sequence, eg starts to list terms of the sequence, with at least 3 correct or \(6n + 3 = 63\) or \((63 - 3) \div 6\ (= 10)\) or \(6 \times 10 + 3\) or \((63 - 27) \div 6\) or \(27 + 6 + 6 + 6 + 6 + 6 + 6\) |
| A1 | eg Yes and 33, 39, 45, 51, 57, 63 Yes and \(n = 10\) Yes and \((63 - 3) \div 6\) is a whole number Yes and \(6 \times 10 + 3 = 63\) Yes and \(63 - 27\) is divisible by 6, and the sequence goes up in 6’s Yes and \(27 + 6 + 6 + 6 + 6 + 6 + 6 = 63\) OR Like all the other terms in the sequence, it is in the 3 times table but not the 6 times table It’s the 10th term / 63 is the 10th term |
Additional guidance
33, 39, 45, 51, 57, 63
‘Yes’ can be implied by an equivalent statement eg ‘63 is in the sequence’
Ignore additional incorrect statements but if the statement contradicts award A0 eg
‘33, 39, 45, 51, 57, 63, and yes and it’s the 11th term’ would be M1A0