Higher November 2022 Paper 3 Q7

(a) Here is the rule for a sequence.

After the first two terms, each term is the sum of the previous two terms

The 1st term is 33

The 2nd term is \(x\)

The 4th term is 73

Work out the value of \(x\). [3 marks]

(b) An expression for the \(n\)th term of a different sequence is \(\quad n - n^2\)

Ruth says,

“All the terms will be negative because \(n^2\) is always greater than \(n\).”

Is she correct?

Tick a box.

Give a reason for your answer. [1 mark]

Mark scheme (a)

\(33 + x = 73\)

Mark scheme (b)

Answer	Mark	Comments
No and gives valid reason	B1	eg No and the first term is zero or No and \(1 - 1^2 = 0\) or No and all the terms are negative except the first

Ignore incorrect or irrelevant statements alongside correct statements
Ignore all other statements and evaluations if \(1 - 1^2 = 0\) seen
Ticks Yes	B0
No and \(0, -2, -6, \ldots\)	B1
No and \(1 - 1^2 = 0\) with \(2 - 1^2 = 1\)	B1
No and \(1 = 1^2\)	B1
No and \(1 - 1 = 0\) (0 is positive) (condone)	B1
No and \(n^2\) can be equal to \(n\) and \(1^2 = 1\)	B1
No and \(n^2\) can be equal to \(n\)	B0
No and \(n\) could equal 1 which cannot become bigger when squared	B1
No and if you put \(n = 1\) it’s not negative	B1
No and \(n = 1\) and \(n^2 = 1\)	B1
No, all the terms are negative except when \(n = 1\)	B1
No and if \(n = 1\) it creates 0	B1
No, not when \(n = 1\)	B0
No, it doesn’t work for the first term	B0
No and \(0.5 - 0.5^2 = 0.25\)	B0
No and when \(n = 0\) it won’t be negative	B0