23 The \(n\)th term of a quadratic sequence is \(\quad an^2 - 5n + c \quad\) where \(a\) and \(c\) are integers.
The first four terms of the sequence are
\(\qquad 2 \qquad x \qquad 16 \qquad y\)
Work out the values of \(x\) and \(y\). [5 marks]
Mark scheme
Answer
Mark
Comments
\(x = 6\) and \(y = 32\)
B5
B4 \(x = 6\) or \(y = 32\) and \(a = 3\) or \(c = 4\) (could be embedded in \(3n^2 - 5n + 4\)) B3 \(a = 3\) or \(c = 4\) (could be embedded in \(3n^2 - 5n + 4\)) B2 any two different correct equations set up for a combination of \(a\), \(c\), \(x\) and \(y\) (may include \(b\). Do not include \(b = -5\)) B1 any correct equation set up for a combination of \(a\), \(c\), \(x\) and \(y\) (may include \(b\). Do not include \(b = -5\)) or (second difference) \(= 18 - 2x\) or \(x + y - 32\) (must be correctly simplified)
Additional guidance
Do not accept any equations with \(n\)
\(x = 6\) on answer line with no working and \(y \ne 32\)
B0
\(x = 6\) and \(y = 32\) on answer line with no working
4 Here is some information about three linear sequences, \(A\), \(B\) and \(C\).
The \(n\)th term of \(C\) is \(42 - 3n\)
The first four terms of \(B\) are
14 22 30 38
\[n\text{th term of } C = n\text{th term of } A + n\text{th term of } B\]
Work out the 20th term of \(A\). [4 marks]
Mark scheme
Answer
Mark
Comments
Alternative method 1: works out \(n\)th term of B
Common difference of 8 or \(8n\)
M1
implied by \(8(n - 1)\)
\(8n + 6\)
M1dep
oe eg \(14 + 8(n - 1)\)
\(42 - 3n -\) their \((8n + 6)\) or \(36 - 11n\)
M1
oe expression for \(n\)th term of A their \(8n + 6\) must be of the form \(an + b\) with \(a\) and \(b\) non-zero missing brackets must be recovered \(36 - 11n\) or \(36 - 11 \times 20\) is M3
\(-184\)
A1
Alternative method 2: works out 20th term of B
Common difference of 8 or \(8n\)
M1
implied by \(8(n - 1)\)
\(14 + 8 \times 19\) or 166
M1dep
oe eg \(38 + 8 \times 16\)
\(42 - 3 \times 20\) or \(-18\)
M1
\(-184\)
A1
Alternative method 3: works out consecutive terms of C
39 36
M1
\(39 - 14\) and \(36 - 22\) or 25 and 14
M1dep
oe
\(25 - 11 \times 19\)
M1dep
oe
\(-184\)
A1
Additional guidance
Up to M3 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
Condone \(-184\) seen in working with answer 184
M3A1
1st M1 Common difference of 8 may be seen by the sequence for B
Alt 3 may be seen using any pair of consecutive terms of C
5 \(X_n\) and \(Y_n\) are the \(n\)th terms of two sequences.
\[X_n = (n - 1)(n + 1)\]\[Y_n = (n + 1)(n + 2)\]
Prove that every term of the sequence with \(n\)th term \(\quad Y_n - X_n \quad\) is a multiple of 3 [3 marks]
Mark scheme
Answer
Mark
Comments
\((n + 1)(.........)\) or \(n^2 + n - n - 1\) or \(n^2 - 1\) or \(n^2 + 2n + n + 2\) or \(n^2 + 3n + 2\)
M1
ie common factor of \((n + 1)\) taken out and attempt to subtract oe
\((n + 1)(n + 2 - (n - 1))\) or \(n^2 + 2n + n + 2 - (n^2 - n + n - 1)\)
M1dep
oe eg could be subtracted vertically
\((n + 1)(n + 2 - n + 1) = 3(n + 1)\) or \(3n + 3\) or \(n^2 + 3n + 2 - n^2 + 1 = 3n + 3\) or \(3(n + 1)\)
A1
no statement needed at end of proof
Additional guidance
\(X_n - Y_n\) can be awarded method marks but A0 unless rescued at the end eg recovery \(-3(n + 1)\) so \(Y_n - X_n\) is a multiple of 3
M2 A0 M2 A1
Missing brackets can be recovered later in the proof (but if not recovered then second M should not be awarded). Must see \(3n + 3\) to show recovery eg \(n^2 + 3n + 3 - n^2 - 1\) (error in \(X_n\) but \(Y_n\) correct with missing bracket)
M2 A1 M1 M0 A0
Condone use of \(x\) (or other letter) instead of \(n\). A mixture of letters would only gain M1. However \(Y_n - X_n = 3(n + 1) = 3x\) would be acceptable to show a multiple of 3
Condone use of a different variable eg \(N\) or \(x\)
Only identifying common difference of 3.5
B0
Mark scheme (b)
Answer
Mark
Comments
\(-6\)
B2
B1 \(318 - 9n \lt 0\) oe inequality or \(318 - 9n \leqslant -1\) oe inequality or \(318 \div 9\) oe calculation or \(319 \div 9\) oe calculation or \(35\dfrac{1}{3}\) or 35.3(…) or \(35\dfrac{4}{9}\) or 35.4(…) or 36(th)
Additional guidance
B1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts
Calculations or values that score B1 may be seen in an equation or an inequality (even if incorrect) eg1 \(n = 106 \div 3\) eg2 \(n \lt 35\dfrac{1}{3}\)
\(3a + b = 1\) and \(a = 2\) substituted in this equation
M1dep
oe
\(b = -5\), \(c = 1\) so \(2n^2 - 5n + 1\)
A1
Additional guidance
Condone other letters used eg \(2x^2 - 5x + 1\) or even \(2n^2 - 5x + 1\)
After finding \(a = 2\) they may find the 0th term to get \(c = 1\)
M2
\(2n^2 + 5n - 1\) from Alt 1 but subtracting the wrong way round
SC2
Notes
(corrected from the printed mark scheme: the fourth equation in Alternative method 2 is printed as \(16a + 9b + c = 13\); for \(n = 4\) it is \(16a + 4b + c = 13\))
Mark scheme (b)
Answer
Mark
Comments
\(n^2 + 10n - 2000 \lt 0\)
M1
the correct inequality needed for this mark and must be written in this form
oe inequality not needed for this mark condone + instead of \(\pm\) as the negative solution has no meaning here
39
A1
Additional guidance
Do not accept T&I
Incorrect use of inequalities can be recovered by a correct use of inequalities later in the method such as \(n \lt 40\) near the end
M2
Incorrect use of inequalities can be recovered for full marks. An answer of 39 after a method that uses an incorrect inequality or = shows inequality has been recovered
M2A1
An incorrect solution with incorrect use of inequalities can only be awarded the second M mark
M0M1A0
Correct answer not coming from correct working will not gain any marks
M0A0
For students who try to complete the square accept \((n + 5)^2 \lt 2025\) as an oe giving M2 but \((n + 5)^2 = 2025\) would only gain M0M1 unless recovered in the answer
Work out an expression for the \(n\)th term. [3 marks]
Mark scheme (a)
Answer
Mark
Comments
\(7 + 12\sqrt{5} + 6(9 - 2\sqrt{5})\) or \(12\sqrt{5} + 6(-2\sqrt{5}) = 0\) or \(12\sqrt{5} \div 2\sqrt{5} = 6\) or states that need to add 6 lots of \((9 - 2\sqrt{5})\) or 7th term
M1
oe eg \(7 + 6 \times 9\) or \(7 + 54\) or \(6 \times -2 = -12\)
oe eg \(2n^2 + 0n - 5\) does not need terms collected
Additional guidance
Condone working in a different variable
Alt 1 2nd M1 Subtracting given terms from \(2n^2\) leading to 5 5 (5 5) must be recovered eg final answer \(2n^2 - 5\) (\(2n^2 - 5n\) or \(2n^2 - 5n - 5\) is not a recovery)
(a) The \(n\)th term of a sequence is \(\quad \dfrac{3 - 5n}{2}\)
Work out the difference between the 20th term and the 8th term. [2 marks]
(b) The \(n\)th term of another sequence is \(\quad \dfrac{3n}{1 - 2n}\)
Write down the limiting value of the sequence as \(n \to \infty\) [1 mark]
Mark scheme (a)
Answer
Mark
Comments
\(\dfrac{3 - 5 \times 20}{2}\) or \(\dfrac{3 - 100}{2}\) or \((-)\dfrac{97}{2}\) or \((-)48.5\) or \(\dfrac{3 - 5 \times 8}{2}\) or \(\dfrac{3 - 40}{2}\) or \((-)\dfrac{37}{2}\) or \((-)18.5\) or \(12 \times (-)\dfrac{5}{2}\)
M1
oe
\((-)30\)
A1
Accept if both 30 and \(-30\) are seen
Mark scheme (b)
Answer
Mark
Comments
\(-\dfrac{3}{2}\) or \(-1\dfrac{1}{2}\) or \(-1.5\)
B1
oe
Additional guidance
Condone \(\dfrac{3}{-2}\) or \(n \to -1.5\) or \(-1\dfrac{1}{2} \to \infty\)
12 Here are the first four terms of linear sequences \(X\) and \(Y\) and quadratic sequence \(Z\).
Sequence \(X\)
7
9
11
13
....
Sequence \(Y\)
2
5
8
11
....
Sequence \(Z\)
14
45
88
143
....
(a) Work out the \(n\)th term of sequence \(X\). [2 marks]
(b) The \(n\)th term of sequence \(Y\) is \(\ 3n - 1\)
Using your answer to part (a), or otherwise, work out the \(n\)th term of sequence \(Z\).
Give your answer in the form \(\ an^2 + bn + c \ \) where \(a\), \(b\) and \(c\) are integers. [3 marks]
Mark scheme (a)
Answer
Mark
Comments
\(2n + 5\)
B2
B1 \(2n + k \quad k \ne 5\)
Additional guidance
Allow any letter eg \(2x + 5\)
B2
Allow \(n =\) eg \(n = 2n + 5\) or \(n = 2x + 5\)
B2
\(2n\)
B1
\(2n + 5 = 0\) is B2 unless also seen with answer of \(-2.5\) which then scores B1
Mark scheme (b)
Answer
Mark
Comments
Alternative method 1
\((3n - 1)\ (\times)\) their \((2n + 5)\)
M1
ft their (a) Brackets needed but may be recovered
\(6n^2 + 15n - 2n - 5\)
M1dep
4 terms with at least 3 correct Implied by \(an^2 + 13n - 5 \ \ (a \ne 0)\) or \(6n^2 + 13n + b \ \ (b \ne 0)\) ft their two term linear expression in (a)
\(6n^2 + 13n - 5\)
A1ft
ft their two term linear expression in (a)
Alternative method 2
(Second differences =) 12 or \(6n^2\) or \(a = 6\)
