Notes

Condone using other letters for a and d (e.g. may use r for d)

M1: Attempts a valid method to solve the problem

• Uses the common difference to form a correct equation and attempts to solve to find a value for \(k\)
  e.g. Attempts to solve \(10 - 6k = 2k - 10\) o.e such as \(2k - 6k = 2(10 - 6k)\)
• Uses 10 as the mean of \(2k\) and \(6k\): \(\dfrac{2k + 6k}{2} = 10 \Rightarrow k = \ldots\)
• Sets up correct equations \(a = 6k,\ a + d = 10\) and \(a + 2d = 2k,\) or may be seen as \(6k + d = 10\) and \(10 + d = 2k\), and proceeds to find \(k\).
• Uses the summation formula \(S_3 = \dfrac{3}{2}(6k + 2k) = 6k + 10 + 2k\) and proceeds to find \(k\)

In each attempt the initial equation (or simultaneous equations) must be correct but do not be concerned by the mechanics of the rearrangement to find \(k\). May be implied by \(k = \dfrac{5}{2}\)

A1: \((k =)\ \dfrac{5}{2}\) o.e.

Notes

Work seen in (a) can only be scored if seen or used in (b)

B1ft: Common difference \(= -5\) or ft on their value for \(k\) (even if \(k\) has been found from an incorrect method) e.g. \(10 - 6 \times \text{``}\dfrac{5}{2}\text{''}\) or e.g. \(2 \times \text{``}\dfrac{5}{2}\text{''} - 10\) if only a numerical value is seen.
May be implied or seen in a term or summation formula.

Note that some candidates may work in terms of \(k\) throughout so only allow B1ft to be scored when they substitute in their numerical value for \(k\), following \(10 - 6k\) o.e. correctly embedded in a correct formula. They may make arithmetical slips before they substitute in their numerical value for \(k\) which can be condoned.

M1: Attempts to use a correct formula. The expression is sufficient to score this mark but they must be using a correct value for \(a\) and \(\pm d\) (or ft on their value for \(k\) for \(a\) and \(d\)) which are correctly placed in the formula.

e.g. \(\left(S_{50} =\right) \dfrac{50}{2}\left(2 \times \text{``}6k\text{''} + 49 \times \text{``}\pm d\text{''}\right)\).

Alternatively, they may find the 50th term \(u_{50} = \text{``}15\text{''} + (50-1) \times \text{``}{-}5\text{''} = -230\) and use \(\left(S_{50} =\right) \dfrac{50}{2}\left(\text{``}15\text{''} + \text{``}{-}230\text{''}\right)\).

If working in terms of \(k\) they must substitute in their value for \(k\)

e.g. \(\left(S_{50} =\right) \dfrac{50}{2}\left(2 \times 6k + 49 \times (10 - 6k)\right) = -7050k + 12250 = -7050 \times \text{``}\dfrac{5}{2}\text{''} + 12250\)

Do not withhold this mark for omission of brackets around \((-5)\)

e.g. \(\dfrac{50}{2}\left(2 \times 15 + (49) - 5\right)\) scores M1

A1: \(-5375\) cao

Notes

Special cases:

SC 1: For those that verify rather than prove a SC 100 is awarded for substituting \(k = \dfrac{5}{2}\) into all three terms to correctly obtain 243, 81 and 27 with a statement that this is geometric with \(r = \dfrac{1}{3}\) (or equivalent reason). All statements must be correct.

SC 2: Be aware that e.g. \(\dfrac{9^{7-2k}}{3^{4k-5}} = \dfrac{3^{2(k-1)}}{9^{7-2k}} \Rightarrow 81^{2(7-2k)} = 9^{6k-7}\) is an incorrect process (without some indication that they have intentionally squared both sides) that fortuitously leads to the correct answer and may score maximum SC 010.

M1: Uses the 3 terms to set up an equation in \(k\) and

• either reaches a common base by replacing 9 with \(3^2\) or by replacing 3 with \(9^{0.5}\) and uses the power law of indices correctly
• or uses the laws of indices correctly to reach \(9^{14-4k} = 3^{6k-7}\) condoning slips in e.g. expanding brackets.

Writing down e.g. \(2(7-2k) - (4k-5) = 2(k-1) - 2(7-2k)\) is sufficient to imply the M1.

dM1: Correct processing leading to a value for \(k\).

A1*: Correct value following correct working. Allow \(k = 2.5\) in place of \(k = \dfrac{5}{2}\)
Condone missing/invisible brackets provided they are recovered correctly.

Alt 1 Using Base 9:

Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.

Alt 2 Finding \(r\) in terms of \(k\) and using e.g. \(u_3 = ar^2\):

Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.

Alt 3 Using Logs Way 1:

Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.

Alt 4 Using Logs Way 2:

Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.

Alt 5 Recognising that taking \(\log_{3}\) forms an Arithmetic Sequence:

Score dM1 when a value of \(k\) is achieved using a correct process and A1* if fully correct.
There is no need to see any mention of log in this approach.

Notes

M1: Deduces expressions for the first term and the common ratio using \(k = \dfrac{5}{2}\) in the correct formulae and finds at least one of \(a = 243\) or \(r = \dfrac{1}{3}\). Allow if seen in (a). May be implied by correct values for \(a\) and \(r\).  For reference, \(a = 3^{4(2.5)-5}\) and \(r = \dfrac{9^{7-2(2.5)}}{3^{4(2.5)-5}}\ \left\{\text{or } r = \dfrac{3^{2(2.5-1)}}{9^{7-2(2.5)}}\right\}\)

M1: Recalls the sum to infinity formula and substitutes their values for \(a\) and \(r\) provided \(|r| \lt 1\)
Dependent on a correct attempt to find both \(a\) and \(r\) using \(k = 2.5\) but allow if neither value is correct or if they are unprocessed e.g. \(\dfrac{3^{4(2.5)-5}}{1 - \dfrac{9^{7-2(2.5)}}{3^{4(2.5)-5}}}\) scores this mark.

A1: cao. Correct sum to infinity. Answer only (with no working) scores full marks. Apply isw.

Notes

M1: Correct application of the given recurrence relation using \(u_1 = 6\) to find \(u_2\) and then \(u_3\) in terms of \(k\) and sets \(u_3 = -1\)
Condone missing brackets if the intention is clear e.g. \(u_2 = 6k - 5 \Rightarrow u_3 = k\,6k - 5 - 5\)

A1*: Obtains the printed answer with no errors including the “= 0”
This is a given answer so do not condone slips/missing brackets unless they are recovered before the final printed answer.

Alternative:

M1: Correct application of the given recurrence relation using \(u_3 = -1\) to find \(u_2\) in terms of \(k\) and then uses \(u_1 = 6\) to find another expression for \(u_2\) in terms of \(k\) and equates the 2 expressions.

A1*: Obtains the printed answer with no errors including the “= 0”
This is a given answer so do not condone slips unless they are recovered before the final printed answer.

Notes

(b)(i)

B1: Deduces the correct value of \(k\). Ignore any working and just look for this value.
Allow equivalent exact values e.g. \(1\frac{1}{3}\) or \(1.\dot{3}\) but not clearly rounded e.g. 1.333
It must be clear that \(k = \dfrac{4}{3}\) is selected so if both roots are offered score B0 unless \(k = \dfrac{4}{3}\) is clearly intended by the calculation in part (ii)

(ii)

M1: Attempts the second term by e.g. \((\text{their } k) \times 6 - 5\) and then adds 6 and \(-1\) to their second term. E.g. \(6 + \text{``}\dfrac{4}{3}\text{''} \times 6 - 5 - 1\)
If they use \(u_1\) and \(u_3\) they must be as given in the question but condone a clear mis-copy of their \(u_2\) value.
The attempt at the second term may be implied by their value.
Note that they may use \(u_3 = -1\) to find \(u_2\) e.g. \(-1 = \text{``}\dfrac{4}{3}\text{''}u_2 - 5 \Rightarrow u_2 = \text{``}\dfrac{3}{4}\text{''}(5-1) = 3\)
Condone slips when rearranging as long as the intention is clear.
The attempt at the second term may be seen embedded in their attempt at the sum e.g.
\(\displaystyle\sum_{r=1}^{3} u_r = 6 + \underline{\frac{4}{3} \times 6 - 5} - 1\) or e.g. \(\displaystyle\sum_{r=1}^{3} u_r = 6 + \underline{\frac{3}{4}(5-1)} - 1\)

If they use both of their values for \(k\) allow M1.

Alternatives:
Note that \(\displaystyle\sum_{r=1}^{3} u_r = 6 + 6k - 5 - 1 = 6k\) so you may just see an attempt at \(6k\) with their \(\dfrac{4}{3}\).
Note that \(\displaystyle\sum_{r=1}^{3} u_r = 6k^2 + k - 4\) so you may just see an attempt at \(6k^2 + k - 4\) with their \(\dfrac{4}{3}\).

A1: Correct value of 8 and no other values unless rejected.
Correct answer with no working scores both marks.
Allow recovery from an inexact value from part (i) e.g. 1.333

Notes

B1*: Correct working to obtain 290. Must be a correct calculation so do not condone missing brackets unless they are recovered. E.g. \(\left(u_{12} =\right) 400 + 12 - 1 \times -10 = 290\) scores B0 unless followed by \(= 400 + 11 \times -10 = 290\). Condone \(\left(u_{12} =\right) 400 + (12-1) - 10 = 290\)
The “£” symbol is not required but the “290” must appear.

Alternative 1:

B1*: Correct working using the 290 to obtain \(n = 12\).
There must be at least one intermediate line after setting up the equation and must be correct work so do not condone missing brackets unless they are recovered (as above).
A conclusion is not required with this approach as long as 12 is correctly obtained.

Alternative 2:

B1*: Correct working using the 290 and 400 to obtain \(d = -10\).
There must be at least one intermediate line after setting up the equation and must be correct work so do not condone missing brackets unless they are recovered (as above).
A conclusion is not required with this approach as long as \(-10\) is correctly obtained.

Allow candidates to list terms and show the 12^th term is 290 e.g.
400, 390, 380, 370, 360, 350, 340, 330, 320, 310, 300, 290
Must list all 12 terms which must be correct and end with 290
Condone if missing 400, 390, 380 as these are given in the question.

Notes

Mark (b) and (c) together

M1: Uses a correct sum formula in terms of \(N\) or \(n\) with \(a = 400\) and \(d = -10\) or \(+10\) and sets = 8100. Condone e.g. > 8100 and allow A1 if this is recovered to become “=” before the final line.
Condone \(8100 = \dfrac{1}{2}N\left(2 \times 400 + (N-1) - 10\right)\) if recovered or not.

A1*: Fully correct proof with sufficient working shown and no unrecovered errors.
Do not condone e.g. missing brackets or e.g. a missing \(N/n\) unless recovered before the final given answer.
Condone the use of \(n\) instead of \(N\) for both marks.
Condone terms in a different order as long as they are correct.
Condone \(0 = N^2 - 81N + 1620\ *\)
Sufficient working requires all brackets to be removed to obtain an unsimplified expanded quadratic before proceeding to the given answer including the “=0”.

Alternative (further maths method): Series summation approach:

M1: Attempt to sum an appropriate series with first term 400. Condone use of \(+10\) as in the main scheme.

A1*: As main scheme.

Notes

M1: Solves the given quadratic equation by any correct method including a calculator to obtain at least one value for \(N\). See general guidance for solving a 3-term quadratic.
If values are just written down and only one value is given it must be 45 or 36.
If both values are just written down they must both be correct.

A1: Realises that the smaller value is required and so selects \((N =)\ 36\).
Ignore any units if given.
The “\(N =\)” is not required, just look for the correct value.
It must be clear that this value has been selected. This may be indicated by e.g. underlining the 36 or the omission of the 45. If the 45 is not rejected score A0.
\(N = 36\) with no working scores M1A1

Typical solution

Typical solution

(b)(i)

(b)(ii)

\(r - r^2\) is at its maximum when \(r = \dfrac{1}{2}\)

Maximum value of \(r - r^2 = \dfrac{1}{4}\)

(b)(iii)

Hence \(T_{\infty} \geqslant 240\)

Typical solution

Typical solution

Typical solution

Convergent

Typical solution

Typical solution

2

Typical solution

Typical solution

Typical solution

Value of top prize = £240

Typical solution

Typical solution

(i)

Total in account = £ 6 787

(ii)

The interest rate is unlikely to remain fixed for the whole 10 years

Typical solution

Typical solution

(i)

(ii)

Typical solution

(i)

(ii)

(iii)

Typical solution

Typical solution

(i)

(ii)

not decreasing

Typical solution

(i)

(ii)

Typical solution

Typical solution

(i)

Ziad will need to train for 29 days

(ii)

Swims 47850 metres

47850 < 50 000

Therefore the coach is not correct

Typical solution

(i)

(ii)

Typical solution

Typical solution

Therefore \(x = 5\) is the only solution

Typical solution

(i)

(ii)

Typical solution

Typical solution

2

Notes

(a)(i)

B1: Must see \(a + 2d\) (either on its own or as one side of an equation) before this answer.

(a)(ii)

M1: May be implied by A1.
May see \(a(a + 3d) = (a + 2d)^2\) oe or \(a\left(\frac{a + 2d}{a}\right)^2 = a + 3d\)

A1: Must be \(d\) in terms of \(a\) so do not accept \(a = -4d\) for this mark.

Alternative method

M1: May be implied by A1.
(NB \(ar - a = 2d\) with (a)(i) substituted is not sufficient)

A1: Must be \(d\) in terms of \(a\) so do not accept \(a = -4d\) for this mark.

Notes

M1: FT their expression from (a)(ii) with \(r = \frac{a + 2d}{a}\) or \(r = \frac{a + 3d}{a + 2d}\). May be implied by A1.

A1: Must see \(\frac{1}{2}\) in their answer for (b) to score these marks.

Notes

(a)(i)

B1: No need to label terms as \(u_2\) etc, and condone 3 appearing at start of list ie 3, 4… would be B1

B1FT: Both correct. FT their \(u_2\). Ignore any further terms given

(a)(ii)

B1: B0 if additional incorrect statement such as ‘diverging AP’ or ‘increasing exponentially’

‘Increasing monotonically’ is fine, but B0 if just ‘monotonic’

Notes

M1: Attempt to find \(r\). Attempts at \(a\) and \(r\) could be in either order, with M1 awarded for a correct method with a possibly incorrect \(a\) or \(r\)

M1: Attempt to find \(a\). Correct \(a\) and/or \(r\) seen in formula for sum to infinity would imply the M mark(s)

A1: Use correct \(a\) and \(r\) to obtain sum to infinity as 54

‘Determine’ so some method needed. Minimum is \(\dfrac{18}{\frac{1}{3}} = 54\).

SC B1 for 54 with no working shown.

Notes

B1: Obtain any correct equation in terms of \(r\) and \(x\)
Could be implied by later work
May use other than \(r\)

M1*: Attempt equation in terms of only \(x\)
This equation would imply the B1
Or correct equation in terms of only \(r\)

A1: Obtain any correct equation not involving fractions or brackets
May still have like terms not yet combined
May result in a cubic depending on method (probably \(6x^3 - 39x^2 - 270x - 297 = 0\))

A1: Solve quadratic BC to obtain both correct \(x\) values
Or solve cubic, to obtain three correct roots (third is likely to be \(x = -1.5\))

A1: Obtain at least \(r = \tfrac{4}{5}\)
If second value of \(r\) given then it must be correct (if third value given then it must be consistent with their correct cubic roots)

M1d*: Attempt sum to infinity, using correct formula, with their \(r\) and their attempt at \(a\)
Must be using their numerical values for \(a\) and \(r\) with \(|r| \lt 1\)
ISW using additional value(s) of \(r\)
M0 if using their \(x\) and not attempt at \(a\)

A1: Obtain 125 only
A0 if additional solution

B1: Clear explanation as to why \(r = -2\) is discarded
Must be considering correct \(r\) value, so B0 if rejecting \(x = -3\) as \(|-3| \gt 1\)
Could generate the terms \(-3, 6, (-12)\) and hence conclude with ‘divergent sequence’
If additional solutions for \(x\) and/or \(r\) from cubic then they must also be correct and explicitly rejected

NB Eliminating \(x\) not \(r\) is a valid method, and could gain full credit. When solving their quadratic there is no need to see \(r = -2\) (and hence \(x = -3\)) to award the A marks

Notes

B1: soi

B1: soi (or \(a_1 + 99d\))

M1: For an attempt at the general term \(b_i\) using \(b_1\ (= a_{100})\), i.e. an expression of the form \(a_1 + kd + (i - 1)(\pm d)\) oe

M1: Equating their \(b_i\) to an expression for \(a_i + 10\)
May see \(a_1 + (i - 1)d + 10\)

A1: www (so this mark is dependent on all previous marks)
May see \(a_1\) used throughout instead of 4
Condone other letters used throughout (e.g. \(n\) for \(i\) and \(c\) for \(-d\)):

• Must recover to \(d\) in order to solve (A1A1)
• Must recover to \(i\) to demonstrate the AG (A1)

Notes

B1: soi (accept e.g. “\(i\) must be an integer”)

B1: Ignore negative values (but B0 for any additional positive values)

Notes

B1: Or 0.5
Must be seen as \(\tfrac{1}{2}\) and not just \(1 - \tfrac{1}{2}\)

B1FT: FT their \(u_2\)
Both as simplified numerical values

Notes

B1: Any correct description, such as repeating.
Condone just ‘periodic’ without the period being stated.

ISW an incorrect period eg ‘periodic with period 4’
B0 if additional incorrect description eg ‘periodic AP’
Allow recurring, repetitive, cyclic etc
Condone looping, circling etc
Do not allow harmonic or alternating, even if with another correct description
B0 for divergent or oscillating, unless additional detail eg between 3 values
Must have a periodic sequence in (a) (with period of at least 3) to gain credit for description

See appendix for further examples (below).

Appendix: exemplar responses for Q4(b)

Notes

M1: Identify that every block of three terms will increase the total by 1.5 (allow use of \(2 + 0.5 - 1\) instead).
Can still award M1 if using the sum of three of their consecutive terms
Must have a periodic sequence in (a) to gain any credit for method (but condone one with a period other than 3)

M1: Attempt to identify the number of terms needed eg 47 blocks plus 2 more terms.
\(73 \div 1.5\) is sufficient for M1 M1
Allow M1 if using blocks of 1.5 to try to find a sum of 73 eg \(48 \times 1.5 = 72\)
Can still get M1 if attempting to use the sum of their three terms

A1: Obtain \(k = 143\)

Notes

M1: Attempt to show reduction to linear form
Introduce logs on both sides, and correctly split to the sum of two terms

A1: Obtain correct equation
Condone no base; any bases seen must be 10
A0 for \(\log_{10}bt\) unless previously seen as \(t\log_{10}b\)

A1: Link to equation of straight line
Could instead refer to linear relationship

If M0 then allow SC B1 for statement such as \(S\) against \(t\) is an exponential function so \(\log S\) against \(t\) will give a straight line

Alternative method

M1: Attempt equation of straight line, and attempt expression for \(S\)
Must be using \(\log_{10}S\) against \(t\)
Must use base of 10

A1: Correctly split into two terms

A1: Link to exponential model

Notes

B1: Link gradient of line of best fit to linear form and confirm \(b \approx 1.15\)
Allow \(m\) in range [0.055, 0.065]
Or \(\log_{10}1.15 = 0.06\) and compare to gradient

B1: Link intercept of line of best fit to linear form and confirm \(a \approx 120\)
Allow \(c\) in range [2.075, 2.085]
Or \(\log_{10}120 = 2.08\) and compare to intercept

Plotted points are linear so may not see line of best fit drawn
If substituting into formula (either given model or linear reduction) then

• B1 for finding and verifying any point that would be on the line of best fit
• B1 for finding and verifying a second point

Notes

M1: Substitute \(t = 7\) into given model
soi

A1: Conclude with integer value
Accept 320 items

Notes

(d)(i)

B1: State or imply sum of GP with \(a\) as 120 or 138, and \(r\) as 1.15
Could be implied by attempt to use GP sum formula (but not just \(n\)^th term) – allow slip as long as clearly sum being considered

M1*: Attempt sum of GP, with \(a = 120\) or 138 and \(r = 1.15\), related to 70000
Must be correct sum formula
May have \(n\) not \(t\) throughout
Allow \(r = 1.15^{t+1}\) with \(a = 120\) but this is B1 M1 only, as not a valid method)

M1dep*: Attempt to rearrange equation as far as \(1.15^t =\)
Must now have \(a = 138\) (or equiv)
Allow sign errors only
Allow T&I as not DR

A1: Obtain 32 (‘months not required’)
www
If 32 given as answer only then allow full marks; if any method shown then mark using the main scheme
Allow BOD with any inequality signs

(d)(ii)

B1: State or imply that the model is unlikely to be valid, with a sensible reason why – could refer to reason why pattern may not continue or extrapolation not being reliable

• Decrease in demand
• Increase in competition
• No values beyond \(t = 6\) so pattern unknown
• Reason why sales are likely to level off / plateau or unlikely to continue to increase (‘other factors’ not enough)
• Item sales may vary according to season

Notes

B1*: Forming a correct expression for \(d\) (or a correct equation containing \(d\)) e.g. \((d =)\,\frac{1}{2}(-\sqrt{3}\cos\theta - 2\sin\theta)\), \((d =)\,\frac{1}{3}\left(\frac{7}{2}\sin\theta - 2\sin\theta\right)(= 0.5\sin\theta)\)
Can be implied e.g. \(\frac{7}{2}\sin\theta = 2\sin\theta + 3(\ldots)\) seen
Can be implied by a correct equation for \(\theta\)

M1dep*: Obtaining an equation of the form \(\tan\theta = k\) from a trigonometric equation which initially had 3 sine and 1 cosine terms or 2 sine and 2 cosine terms e.g. if correct \(\frac{7}{2}\sin\theta = 2\sin\theta + 3\left(\frac{7}{2}\sin\theta + \sqrt{3}\cos\theta\right)\)

A1: Condone \(-\dfrac{\pi}{6}\) stated too but A0 if any other value given in the interval \(\frac{1}{2}\pi \lt \theta \lt \pi\) (but ignore any values that are given outside this range)
Exact answer must be seen at some stage

Notes

B1ft: Correct formula for the sum of an AP with \(a = 2\sin\theta\) (with either \(\theta\) or their value of \(\theta\) substituted) and either \(d\) or their value of \(d\) substituted or their expression for \(d\)
Follow through their values of \(\theta\) and \(d\) if used provided \(\frac{100}{2}[2(2\sin\theta) + (100 - 1)d]\) implied

B1ft: Correct expression for \(d\) using their \(\theta\) (e.g. \(d = \frac{1}{2}(-\sqrt{3}\cos\theta - 2\sin\theta)\), \(d = \frac{1}{3}\left(\frac{7}{2}\sin\theta - 2\sin\theta\right)\))
Follow through their value of \(\theta\) only

B1: www – must have come from \(\theta = \frac{5}{6}\pi\) correctly derived in (a) oe (not for 1338 or 1340 unless 1337.5 seen so isw once 1337.5 (oe e.g. \(\frac{2675}{2}\)) seen)
Correct answer with no working scores all 3 marks

Notes

M1: Sum of a GP implied. Allow any \(r\), eg \(r = 0.05\)

A1: Correct equation

A1: Allow 0.487 or 0.488

M1: or \(0.95^{14} = 0.4875\) or 0.487 or 0.488 seen. Can be implied by their answer
ft their equation of form \(a^n = b\) (dep M1 gained and \(b \gt 0\))

A1: cao. Allow \(n = 14\). Allow 14.0. Allow \(\approx 14\)

Alternative method

M1: Attempt add \(\geqslant 10\) terms. Allow any value of \(r\) for this mark

A1: Correct 14 terms added

NB Unsupported correct answer: SC B3

Alternative (incorrect) methods using \(r = 1.05\), or \(\frac{1}{0.95}\) or \(\frac{20}{19}\)

(For info’ only: \(r = \dfrac{1}{0.95}\) or \(\dfrac{20}{19}\) comes from misinterpreting “lowest” to mean “shortest”)

M1: Allow any value of \(r\) for this mark

A1: oe using 1.05. Correct equation

M1: oe, eg \(\dfrac{\ln 1.539}{\ln 1.053}\) or \(\dfrac{\ln 1.51}{\ln 1.05}\)   ft their “\(\frac{117}{76}\)”
Can be implied by their answer

Notes

M1*: M1 for un-simplified correct use of \(r = \dfrac{u_n}{u_{n-1}}\) so allow this mark for stating that \(r = \dfrac{15}{xy}\) or \(r = \dfrac{yz}{15}\) or \(r^2 x = z\) oe
Or for the terms of the GP as \(y + 4, \dfrac{15}{y}, y - 4\) in \(y\) only

M1*: M1 for \(y - x = \pm 4\) or \(z - y = \pm 4\) or \(z - x = \pm 8\) oe
Or for the terms of the AP as \(y + 4, y, y - 4\) in \(y\) only
For reference: \(xy^2z = 225\), \(x = y + 4\) and \(z = y - 4\)

M1dep*: Eliminate \(x\) and \(z\) to form an equation in \(y\) only (must be equivalent to a quartic in \(y\))
Or for \(\dfrac{\;y - 4\;}{\frac{15}{y}} = \dfrac{\frac{15}{y}}{y + 4}\) \(\left(\Rightarrow y^2 - 16 = \dfrac{225}{y^2}\right)\)

A1: AG so sufficient working must be shown www – note that \(y = x + 4\) and \(y = z - 4\) (from \(d = +4\)) can lead to the correct equation (which can score the M marks only)

Notes

B1: BC
Allow implied e.g. \(y = \pm 5\) then using \(y = 5\) only (B0 if the sum to infinity for other values of \(y\) are not rejected)
\(x = 9\), \(z = 1\)

M1: Calculate \(r\) (soi) corresponding to \(y = 5\)
allow unsimplified e.g. \(r = \dfrac{(5)(1)}{15}\) and allow if more than one value of \(y\) stated
Possibly done implicitly in formula for \(S_\infty\)

M1: Using the correct formula for the sum to infinity of a GP with their value of \(x\) (= their \(y \pm 4\)) and a value of \(r\) where \(-1 \lt r \lt 1\)

A1: cao oe eg \(\dfrac{27}{2}\) - do not award this mark if more than one value for \(S_\infty\) stated
A0 for a triple-decker fraction e.g. \(\dfrac{9}{\frac{2}{3}}\)

Notes

M1 for one error, eg \(a + 9d = 88\)

M1 for \(\frac{15}{2}(2a + 14d) = 310\) or for one error, eg \(5(2a + 18d) = 310\)

M1: Attempt substitution, elimination or to verify solutions found BC.
Condone one arithmetical error
SC for last 2 marks: Correct answers, substitution not seen: B1B0, dep M1M1

A1: cao

Notes

B1: Or for \(a + 2d = ar\)

B1: Or for \(a + 12d = ar^2\)

M1*: Setting up an equation in \(d\) or \(r\) only – dependent on one B mark
\(2 + 12(r - 1) = 2r^2\)

M1dep*: Solving their two-term quadratic equation in \(d\) (or three-term quadratic in \(r\))
\(r^2 - 6r + 5 = 0\)
\((r - 5)(r - 1) = 0\)
\(\Rightarrow r = \ldots\)

A1: Correct value for \(d\) and link to increasing sequence – must either say that \(d\) is positive (oe) or state at least the correct first four terms and comment that they are increasing
Condone no mention of \(d \neq 0\)

Notes

B1: or for \(\dfrac{u_3}{u_1}\)

M1*: Setting up an equation in \(d\) only – dependent on one B mark

M1dep*: Solving their two-term quadratic equation in \(d\)

A1: As above

Notes

M1: Finding common denominator or attempt to rationalise denominator for at least one fraction e.g. \(\frac{1}{\sqrt{n+1}-1}\cdot\frac{\sqrt{n+1}+1}{\sqrt{n+1}+1} - \frac{1}{\sqrt{n+1}+1}\cdot\frac{\sqrt{n+1}-1}{\sqrt{n+1}-1}\) or \(\frac{\sqrt{n+1}+1}{n} - \frac{\left(\sqrt{n+1}-1\right)}{n}\) oe
Condone consistently considering any general term in the sequence e.g. \(a_{k+1}\) for all marks \(\dfrac{1}{\sqrt{k+2}-1}\cdot\dfrac{\sqrt{k+2}+1}{\sqrt{k+2}+1} - \dfrac{1}{\sqrt{k+2}+1}\cdot\dfrac{\sqrt{k+2}-1}{\sqrt{k+2}-1}\)

A1: Correct simplification
\(a_{k+1}\) leads to \(\frac{2}{k+1}\)

A1: nfww
Correct explanation of why the general term must be rational e.g. in form \(\frac{p}{q}\) oe \(p\), \(q\) integers
Condone rational as \(n\) an integer or \(k\) integer
‘\(\frac{2}{n}\) is rational’ alone is A0 - must include comment that \(n\) is an integer

Notes

M1: Must relate to \(\frac{2}{n}\), \(a_n \to 0\) is not sufficient
Allow "\((n > 0\) and\()\ \lim\limits_{n\to\infty}\frac{2}{n} = 0\)"
Condone omission of \(n > 0\)
May find the difference between 2 terms or use inequalities
Allow \(\mathrm{f}(x) = \frac{2}{x} \Rightarrow \mathrm{f}^\prime(x) = -\frac{2}{x^2} < 0\) (condone \(n\) instead of \(x\))
Condone listing terms for M1 A0
FT their \(\frac{2}{n}\) provided in the form \(\frac{b}{n+c}\) where \(b, c\) integers

A1: See appendix
Convincing conclusion e.g. difference between terms is negative, so sequence is decreasing or \(\frac{2}{n+1} < \frac{2}{n}\), so decreasing or function \(\mathrm{f}(x)\) is decreasing therefore sequence \(a_n\) decreasing
A1 is only available if their \(\frac{2}{n}\) is correct for their term e.g. \(\dfrac{2}{k+1}\) from considering \(a_{k+1}\)
SC B1 for ‘decreasing’ following M0 for a correct but incomplete argument. But not just for ‘decreasing’ unsupported.

Appendix: exemplar responses for Q11(b)

Notes

M1: or identify AP with \(n\) terms and \(a = 1\), \(d = 2\); may be implied by award of first A1

A1: use of formula for sum of arithmetic progression with \(n\) terms and \(a = 1\) and \(l = 2n - 1\); or \(a = 1\) and \(d = 2\); allow \(\frac{n}{2}(2 + 2n - 2)\) but no further contraction

A1: correct intermediate step needed for A1

Alternative method (Further Maths 1)

M1: from \(\sum_{r=1}^{n}(2r-1)\); allow for \(\sum_{r=1}^{n}(2r+1)\); may be implied by award of first A1

A1: all correct;

A1: correct intermediate step needed for A1

Alternative method (Further Maths 2)

A1: correct intermediate step and statement needed

Notes

B1: Forming a correct equation
Or first 4 four terms seen \(6, 2, \frac{2}{3}, \frac{2}{9}\)
soi by correct answer or correct value for \(r\)

M1: Attempt to solve their equation for 4^th term of a geometric sequence including taking correct root for their equation
Their equation must be from an attempt to give 4^th term of a geometric sequence e.g. \(6r^4 = \frac{2}{9} \to r^4 = \frac{1}{27}\) so \(r = \sqrt[4]{\frac{1}{27}} = 0.438[6\ldots]\)
May be implied by correct answer
Ignore extra values of \(r\) for M1

A1: cao, nfww
Uses \(\frac{a}{1-r}\) to find \(S_\infty = \dfrac{6}{2/3}\), answer must be simplified.
A0 if other values also found unless explicitly rejected

Notes

B1: Must be in lowest terms. Allow \(2\frac{1}{12}\) BC

Notes

M1: Correct function to generate odd numbers. Condone no brackets

A1: Correct limits for their expression using sigma notation
Brackets must be seen.
Do not allow \(\sum_{r=1}^{5}(r+(r-1))\), \(\sum_{r=1}^{5}(2n-1)\) or similar

Notes

B1: Explains in technical language or otherwise that the sum gets bigger the more terms added in so the sum to infinity is not a number. See appendix

Appendix: exemplar responses for Q3(c)

Notes

M1: Allow instead for \(r = \frac{k+2}{3k-6}\) or \(r^2 = \frac{k+2}{5k-2}\) in any form Soi

M1: Forms an equation in \(k\) which need not be simplified

A1: AG Rearranges to correct three term quadratic.
At least one intermediate step must be shown.

SC1: for showing \(k = 1\) leads to \(3, -3, 3\) \((r = -1)\) and that \(k = 10\) leads to 48, 24, 12 \(\left(r = \frac{1}{2}\right)\) and demonstrating that both are geometric

Notes

M1: Solves the quadratic to give at least one root Soi

M1: Evaluating the terms of the sequence when \(k = 1\)

A1: cao

Alternative for the last 2 marks

M1: Using the formula for the sum of terms of a GP with \(r = -1\)

A1: cao

Notes

B1: Identifies the first term and common ratio soi

B1: cao

Notes

B1: B0 if wrongly attributed

Notes

E1: allow
divergent because ratio of consecutive terms is tending to 1 not 0;
divergent because the terms (in the sequence) are not tending to a finite limit oe
divergent because terms tend to infinity oe

do not allow
divergent because not convergent oe
divergent because terms (in the sequence) increase infinitely oe
divergent because terms get bigger oe

Notes

M1: eg may see \(3 + (29-1) \times 5\); \(a\) must be \(u_0, u_1, u_2, u_3\) or \(u_4\) in the AP and \(d\) must be 5; allow correct full list of terms from 3 to 138 for M1
must see at least eg \(-2 + 29 \times 5\)

A1: if M0 allow SCB1 for 143 not fully supported

Notes

M1: \(a\) must be \(-2\) and \(d\) must be 5; allow sum of full list of terms from \(-2\) to 143 for M1; allow if correct full list seen in part (c) only
must see at least \(15 \times (-2 + \text{their } 143)\) or \(15 \times (-4 + 29 \times 5)\) for M1

A1: if M0 allow SCB1 for 2115 not fully supported

Notes

B1: 2.45 (need not have working)
May be done BC

B1: Convincing completion

Notes

M1: uses laws of logs

A1: AG \(\log\frac{200}{20}\) or \(\log_{10}10\) must be seen explicitly

Alternative method

M1: Uses laws of logs

A1: Must see \(\log_{10}100 = 2\) or \(\log_{10}10 = 1\) explicitly

Notes

M1: Attempts to establish a common difference of 1

A1: argues from a common difference eg “increases by 1 each time”. Must use exact values to establish the difference between the second and third terms.

Alternative method

M1: rewrites two more terms of the sequence and makes a comparison

A1: argues from a common difference eg “increases by 1 each time” Must use exact values to establish the difference between the second and third terms.

Notes

M1: Uses the formula with first term \(\log_{10}20\) and common difference 1

A1: Allow for fully correct expression not simplified. isw
Correct forms include \(50\log_{10}20 + 1225\), \(25\log 400 + 1225\), \(25(\log 400 + 49)\), \(25(\log 4 + 51)\) etc

Notes

B1: Substitute correctly into formula
Condone \(r\) but not \(n\) instead of \(x\)

B1: Integral correctly evaluated

B1: Correct single fraction – may not be fully simplified
OR \(\dfrac{n(n+1)(2n+1)}{6} = \dfrac{2n^3 + 3n^2 + n}{6}\)

B1: Correct completion
Intermediate step needed
Dep on B3
OR convincing comparison of \(\dfrac{n(n+1)(2n+1)}{6}\) and \(\dfrac{n^3 - 1}{3} + \dfrac{6n^2 + 2n + 4}{12}\)

Notes

B1: Method need not be shown
May be done BC

Notes

M1: Correct substitution into formula
Condone \(r\) but not \(n\) instead of \(x\)
Square numbers may be evaluated

A1: Correct completion.
Integration may be done BC
At least one step to be shown

Notes

M1: \(a\) and \(r\) are numerical values; \(a = 9\) and/or \(r = \pm\frac{1}{3}\)

A1: if unsupported allow SC2 for correct answer

Notes

B1: Cao
Must be clear that the limit is equal to 1. Therefore tending to or approaching 1 or 0.999 or \(\to 1\) etc do not score

Notes

M1: Convincing explanation that \(\dfrac{1}{n+1}\) decreases

E1: Convincing completion (A.G.)

Special Case (Max 2 for this method)
Correct differentiation and states \(> 0\) to demonstrate increasing SC B1
Goes on to consider [positive] integers as a subset of reals to complete their argument SC B1 dep on first B1.

Alternative method 1

M1: This method is testing \(\dfrac{n}{n+1} < \dfrac{n+1}{n+2}\)
Formulating expression for \((n+1)\)th or \((n-1)\)th term

M1: For formulating inequality and cross multiplying

E1: Probably see \(\dfrac{n+1}{n+2} > \dfrac{n}{n+1}\) [i.e. increasing]

Alternative method 2

M1: Formulating expression for \((n+1)\)th term

M1: Finding difference and attempts to form a single fraction eg may be \(a_n - a_{n+1}\)

E1: Convincing completion (A.G.)

Notes

DR: This question included the instruction: In this question you must show detailed reasoning.

B1: Substituting values

M1: Attempt to rationalise denominator for one term
Either \(\times\dfrac{\sqrt{2}-1}{\sqrt{2}-1}\) or \(\dfrac{\sqrt{2}-1}{2-1}\) at least once for M1

A1: All correct

A1: Convincing completion (AG)

Notes

(i) B1: Do not allow repeating, recurring etc.
The sequence can also be described as oscillating

(ii) B1: cao

Notes

(i) B1: Allow for one correct inequality

B1: Must have “or” or the union of sets
Condone \(b < -1\) or \(|b| > 1\)

Note for \(b = 1\), the sequence is convergent, but the corresponding series is divergent

(ii) M1: Using the sum of geometric series with \(r = \frac{1}{3}\)

A1: www

Notes

B1: NB \(17 + (n-1) \times 11 = 292\)

Notes

M1: \(\dfrac{\text{their } 26}{2} \times (2 \times 17 + (\text{their } 26 - 1) \times 11)\)

Notes

M1: Use of GP with common ratio 0.98
GP implied by formula for \(S_\infty\), \(S_n\) or \(u_n\) showing 0.98 or 1^st 3 terms

M1: Sum to infinity

Notes

M1: At least two terms written out, allow one error, OR correct answer
Terms may not be seen as may be done completely on calculator