Calculate the value of the common ratio.

Calculate the 10^th term of this sequence.

The k^th term is the first term which is greater than 2000. Find the value of k.

Write in words \((q \wedge \neg r) \Rightarrow \neg p\).

Consider the statement “If the number is a multiple of five, and is not even then it will not end in zero”.

Write this statement in symbolic form.

Consider the statement “If the number is a multiple of five, and is not even then it will not end in zero”.

Write the contrapositive of this statement in symbolic form.

u₁r⁴ = 324     (A1)

u₁r = 12     (A1)

r³ = 27     (M1)

r = 3     (A1)(G3)

Note: Award at most (G3) for trial and error.

[4 marks]

4 × 3⁹ = 78732 or 12 × 3⁸ = 78732     (A1)(M1)(A1)(ft)(G3)

Note: Award (A1) for u₁ = 4 if n = 9 , or u₁ = 12 if n = 8, (M1) for correctly substituted formula.

(ft) from their (a).

[3 marks]

4 × 3^k−1 > 2000     (M1)

Note: Award (M1) for correct substitution in correct formula. Accept an equation.

k > 6     (A1)

k = 7     (A1)(ft)(G2)

Notes: If second line not seen award (A2) for correct answer. (ft) from their (a).

Accept a list, must see at least 3 terms including the 6^th and 7^th.

Note: If arithmetic sequence formula is used consistently in parts (a), (b) and (c), award (A0)(A0)(M0)(A0) for (a) and (ft) for parts (b) and (c).

[3 marks]

If the number is even and the number does not end in zero, (then) the number is not a multiple of five.     (A1)(A1)(A1)

Note: Award (A1) for “if…(then)”, (A1) for “the number is even and the number does not end in zero”, (A1) for the number is not a multiple of 5.

[3 marks]

\((p \wedge \neg q) \Rightarrow \neg r\)     (A1)(A1)(A1)(A1)

(A1) for \(\Rightarrow\), (A1) for \(\wedge\), (A1) for p and \(\neg q\), (A1) for \(\neg r\)

Note: If parentheses not present award at most (A1)(A1)(A1)(A0).

[4 marks]

\(r \Rightarrow (\neg p \vee q)\)   OR   \(r \Rightarrow \neg (p \wedge \neg q)\)     (A1)(ft)(A1)(ft)

Note: Award (A1)(ft) for reversing the order, (A1) for negating the statements on both sides.

If parentheses not present award at most (A1)(ft)(A0).

Do not penalise twice for missing parentheses in (i) and (ii).

[2 marks]

An easy ratio to find and the majority of candidates found r = 3, though many had trouble showing the appropriate method, thus losing marks.

A fairly straightforward part for most candidates.

The majority found k − 7; many without supporting work which lost them a mark. Where candidates had difficulty in this part, it was generally a case of poor algebraic skills.

This question on logic was straightforward for most candidates who scored full marks for parts (a) and (b) (i). A few omitted the brackets in part (b).

This question on logic was straightforward for most candidates who scored full marks for parts (a) and (b) (i). A few omitted the brackets in part (b).

Very poorly answered with many candidates scoring just one mark. The main error was to open the bracket and not use the “or”.