Foundation June 2018 Paper 3 Q14

OCRFoundationCurrent spec7 marksSequences

14 Here are the first four terms of a sequence.

\[6 \qquad 10 \qquad 14 \qquad 18\]

(a) Write down the next term. [1]

(b) Write an expression for the \(n\)th term. [2]

(d) Find the term in the sequence that is nearest to 511. [3]

Answer	Marks	Part marks and guidance
22	1		Condone extra correct terms beyond 22

Answer	Marks	Part marks and guidance
\(4n + 2\) oe	2	B1 for \(4n\)	oe may be a form of \(6 + 4(n - 1)\)

Mark scheme (c)

Answer	Marks	Part marks and guidance
The numbers are even oe or 511 is odd oe or \(n\) is [a] decimal	1	Even must clearly refer to the terms Odd must refer to 511 (or “it”)	See Appendix Only after their \(4n + 2 = 511\) solved to a decimal (may be 127.25)

Exemplars: 14c

Comment	Mark
It isn’t an even number	1
It’s not a multiple of 2 Same as “It is odd”	1
It’s an odd number In this context, “it” clearly refers to 511	1
Because 127 \(\times\) 4 + 2 = 510 and 128 \(\times\) 4 + 2 = 514 and there is nothing between.	1
The sequence goes up in 4s and 11 isn’t a multiple of 4 True but ignores the constant in \(4n + 2\)	0
4 doesn’t go into 511	0
It goes up in the 4 times table and numbers in the 4 times table are even Recognises “even” but for the four times table and not the sequence.	0

Mark scheme (d)

Answer	Marks	Part marks and guidance
510	3	B2 for (127 or 514) as answer OR M1 for their \(4n + 2 = 511\) or better soi 127.25 M1 for their 127.25 (rot) correctly substituted in their \(4n + 2\) OR M2 for trials using their \(4n + 2\) leading to 509 < integer terms < 513 or M1 for two correct trials using their \(4n + 2\) If 0 scored SC1 for 128 as answer	FT for method only if their (b) is of form \(an + b\) where \(a \ne 0\) and \(b \ne 0\) Look back to 14b Rounded or truncated A trial is substituting a value for \(n\) in their \(4n + 2\) (allow adding their 4 after first calculation) May be 22, 26, 30... 506, 510, 514 May be 22, 26, 30... but with errors