P4.5 Geometric series

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (a) was a discriminatory question with many different approaches seen; the mix of algebraic indices coupled with a geometric sequence caused problems for many candidates and often work lacked structure. It was evident that many candidates are not as familiar with the laws of indices as they should be and are not practised in the set-up of proof questions. The layout of candidates work varied enormously and they should be encouraged to practice showing a proof that flows and that is not interrupted by their thought processes. Despite the challenges many faced in part (a), the majority of candidates found part (b) to be accessible due to the given answer in (a) providing a restart opportunity. In part (a), many candidates were able to achieve full marks by setting up an appropriate equation, changing to the common base of 3 and using the index laws to form a linear equation in k and then solving. The most common (and successful) method was using 2 ar ar a ar = and then changing to the common base of 3, using the power law of indices and, either subtracting the indices or by cross-multiplying and then adding the indices. The most common error, which was highly prevalent, was incorrect use of the division/subtraction law of indices – whereby they divided indices rather than subtracted and then formed quadratic equations in k, resulting in no further marks being awarded. Most initial difficulties arose from not understanding how to use the power law correctly for 7 2 9 k - or failing to notice that the bases needed changing at all and many simply subtracted the indices regardless of the differing bases. Changing the base to 9 was rarely seen and attempts at using logarithms were mostly unsuccessful as again, the laws were not generally adhered to. Some candidates demonstrated a lack of understanding of indices by multiplying the bases but adding the indices to achieve ( ) 2 7 2 6 7 81 9 k k - - = . These were given some benefit of doubt and awarded one mark by a special case in the mark scheme, provided the rest of their processing was correct. Other candidates tried to verify the sequence being geometric by substituting in the given value of k, which did not satisfy the demand of the question. These attempts were allowed to score a maximum of 1 mark but were often unsuccessful as not enough of a conclusion was made. It is also worth noting that candidates should pay attention to the form of the given value and realise that this is what is required; an equivalent fraction 35 14 k = was not accepted for the final mark.

Common mistakes

• equate ratios (u₂² = u₁u₃)
• write 9 as 3² and use index laws to get a linear equation in k
• index-law errors very common
• verifying k = 5/2 earns only a special case

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Part (b) was well attempted and generally candidates were highly successful even if part (a) was poorly attempted or not attempted at all. The majority of candidates were able to evaluate both a and r having used k to find the terms in the sequence first, they then used the sum to infinity formula correctly to achieve the answer, resulting in 3 marks. Unfortunately, some candidates found the reciprocal of the ratio which gave them a value of 3 and some mistakenly thought that 5 2 r = , neither of these values met the limitations of the infinite sum formula but this was not spotted by candidates.

Common mistakes

• a = 243, r = ⅓, S∞ = a/(1 − r) = 729/2
• r = 3 (reciprocal) or 5/2 is invalid since modulus of r must be < 1

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was a more demanding than expected question on geometric sequences with a number of candidates getting confused with the value of k and the subsequent values of a and r required for the question. There was still access to later parts as the quadratic in k was given such that part (b) and part (c) could be attempted without part (a). Most candidates were able to score full marks in part (a) by correctly giving an equation in k, almost always 12 3 16 3 4 12 3 k k k k - + = + - and proceeding to the given quadratic with no errors seen. A noticeable number of candidates failed to gain the final mark as they gave a 40 - term in their quadratic. Given that this was a printed answer, it was disappointing that candidates did not identify their error and correct it. Some candidates wrote down a correct expression for r but failed to create an equation, whilst a few candidates formed an equation for the sum of three terms, but invariably failed to expand and simplify correctly. There were several valid but more complicated efforts such as expressing the third term as the first term multiplied by the square of the ratio. This was more likely to result in errors and there were a significant number who did not know where to start. The common mistake where they could not set up an equation in r was to look for a common difference as if it was an arithmetic sequence.

Common mistakes

• equate the common ratios (or use u₂² = u₁u₃)
• not a common difference

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was a more demanding than expected question on geometric sequences with a number of candidates getting confused with the value of k and the subsequent values of a and r required for the question. There was still access to later parts as the quadratic in k was given such that part (b) and part (c) could be attempted without part (a). In part (b)(i) candidates were told that the given sequence converges. Candidates were required to find the value of k and to give a reason for their answer. Solving from calculator, or without working, was allowed and the majority of candidates were able to correctly solve the quadratic reaching the terms 20 k = and 2 3 k = . However, many found giving a reason particularly challenging. Of those who did score both marks, listing values in the sequence was a common explanation, as was finding the value of r and using the fact that 1 r . When attempting to reason via a calculation of the common ratio, there were often errors. A common response was to assume k was equal to r and then selecting 2 3 k = as the final value. Some did not select a final value, keeping 20 and 2 3 . Some candidates who had chosen 20 k = gave comments such as “because it is an integer/whole number” or “because it is a constant”. A significant number chose 20 k = giving a reason “because the sequence is convergent” which they had already been told and did not demonstrate an understanding of what the term means. Part (b)(ii) proved to be very challenging for many candidates. It was quite often the case that candidates tried to use 2 3 r = and just found the value 6 a = when 2 3 k = . It was very frequent that those who had correctly identified 20 k = in part (b)(i) did not show their working to find a and r and lost all marks in (b)(ii) as the sign error in substituting in 3 4 - meant that it was unclear whether the correct value for r had been found, or if the correct sum to infinity formula was being used. Some candidates having found 256 7 correctly then proceeded to give a rounded decimal as their answer which was condoned. Those who found 6 a = and 5 3 r = were still able to score the B mark for a correct pair of values for a and r, even though they could not score the mark for using the sum to infinity formula with an invalid value for r.

Common mistakes

• k = 20 because the modulus of r (¾) is less than 1 (k = ⅔ gives r = 5/3)
• S∞ = a/(1 − r) with a = 64, r = −¾

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The majority of candidates found this question accessible and were able to score consistently throughout. Where marks were lost they were usually in parts (b) and (c). 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐= 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 6𝑡𝑡𝑎𝑎𝑎𝑎𝑎𝑎 Part (a) was attempted successfully by most candidates, starting usually with 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. Candidates then successfully cross multiplied and used the was 6𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡= 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቀ 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቁ 2 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 appropriate identity to reach the required result. Another starting point regularly seen . Errors in the proof were rare if candidates started from a correct statement. Occasionally the “ = 0” was carelessly missing. 𝜋𝜋 In part (b) most candidates correctly solved the equation either by factorising or using 6 rather than the required 5𝜋𝜋 5𝜋𝜋 calculator technology, but many reached the acute angle 6 . Others found a number of solutions but failed to identify 6 as the obtuse angle of 𝜋𝜋 required answer. A few candidates worked in degrees and again lost the accuracy mark. 6 in (b) then usually carried this error into (c) and 𝜋𝜋 Those who had obtained an angle of 6 in (c) often would reach 9൫1 + sqrt 3൯, but did not seem to worry that this did not match the required answer were in a position only to earn the method marks. Candidates using format given in the question. There were many attempts, however, to use the correct formula for the sum to infinity with appropriate values for the first term and the (often incorrectly) algebraic expressions for the sum to infinity involving ��𝜃, without common ratio. Some candidates unfortunately put effort into finding and rearranging 𝑎𝑎 1+𝑟𝑟 incorrectly as the formula substituting an angle in, gaining no credit, and others used for the sum to infinity. The most common mistake in (c) was failure to show the steps 3 ൰ straight into their calculators to obtain 9 -9 sqrt 3 losing the last two marks in -6 sqrt 3 where the denominator was rationalised. Many candidates clearly typed an expression 1-൬- sqrt 3 like the question.

Common mistakes

• equate the common ratios (5 + 2sinθ)/12cosθ = 6tanθ/(5 + 2sinθ)
• tanθ × cosθ = sinθ
• "= 0" needed

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

The majority of candidates found this question accessible and were able to score consistently throughout. Where marks were lost they were usually in parts (b) and (c). 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐= 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 6𝑡𝑡𝑎𝑎𝑎𝑎𝑎𝑎 Part (a) was attempted successfully by most candidates, starting usually with 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. Candidates then successfully cross multiplied and used the was 6𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡= 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቀ 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቁ 2 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 appropriate identity to reach the required result. Another starting point regularly seen . Errors in the proof were rare if candidates started from a correct statement. Occasionally the “ = 0” was carelessly missing. 𝜋𝜋 In part (b) most candidates correctly solved the equation either by factorising or using 6 rather than the required 5𝜋𝜋 5𝜋𝜋 calculator technology, but many reached the acute angle 6 . Others found a number of solutions but failed to identify 6 as the obtuse angle of 𝜋𝜋 required answer. A few candidates worked in degrees and again lost the accuracy mark. 6 in (b) then usually carried this error into (c) and 𝜋𝜋 Those who had obtained an angle of 6 in (c) often would reach 9൫1 + sqrt 3൯, but did not seem to worry that this did not match the required answer were in a position only to earn the method marks. Candidates using format given in the question. There were many attempts, however, to use the correct formula for the sum to infinity with appropriate values for the first term and the (often incorrectly) algebraic expressions for the sum to infinity involving 𝜃𝜃, without common ratio. Some candidates unfortunately put effort into finding and rearranging 𝑎𝑎 1+𝑟𝑟 incorrectly as the formula substituting an angle in, gaining no credit, and others used for the sum to infinity. The most common mistake in (c) was failure to show the steps 3 ൰ straight into their calculators to obtain 9 -9 sqrt 3 losing the last two marks in -6 sqrt 3 where the denominator was rationalised. Many candidates clearly typed an expression 1-൬- sqrt 3 like the question.

Common mistakes

• a = −6√3, r = −1/√3 (modulus less than 1)
• S∞ = a/(1 − r) and rationalise the denominator to 9(1 − √3)
• using θ = π/6 gives 9(1 + √3), which does not match the given form
• steps must be shown, not calculator

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was yet another good source of marks with the modal mark being 6 out of 6, achieved by over 35% of the candidature. Almost every candidate gained the B1 mark in part (a) with a fairly even split between those who used 20 000 x 1.082 and those who first calculated the profit after Year 1 = 21600 and then Year 2 = 23328 showing each step of their working to gain the mark.

Common mistakes

• 20 000 × 1.08² = 23 328 with the method shown
• stating Year 2 and Year 3 values alone is not enough

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was yet another good source of marks with the modal mark being 6 out of 6, achieved by over 35% of the candidature. Part (b) was more problematic for many candidates. There was a very good understanding of the inequality/equation to be formed, and the use of logs was almost always very good. However, a significant minority of candidates failed to gain the final A mark giving an answer of 15.3 or 16 instead of Year 17.

Common mistakes

• 20 000 × 1.08ⁿ⁻¹ > 65 000
• solve with logs: n − 1 > ln 3.25 / ln 1.08 = 15.3
• answer Year 17, not 15.3 or 16
• a sum formula is wrong

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was yet another good source of marks with the modal mark being 6 out of 6, achieved by over 35% of the candidature. Part (c) proved to be very well answered with the vast majority using the sum formula correctly. Errors included using n =19 instead of 20 in the formula or finding the 20th term instead of the sum. A few candidates listed all the terms before adding them which often led to accuracy errors.

Common mistakes

• S₂₀ = 20 000(1.08²⁰ − 1)/(1.08 − 1) = £915 000
• n = 19 or finding the 20th term instead of the sum
• listing terms causes accuracy errors

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question was fairly well answered by a number of candidates, who understood the need to use the sum to infinity formula, but a sizeable majority made little or no progress, not realising for example, that the sum started from n = 2 and not n =1 – hence they often did not get the first mark for a correct first term and/or common ratio. 9/16 was seen fairly often, but not always attributed to the first term, with some candidates thinking this was the value of the sum. One candidate misread the infinity symbol as an 8 and applied the sum to 8 terms correctly, but was obviously not able to score any marks. A number of students found the sum to infinity, using n = 1 to infinity and then subtracting the first term. This was done well by candidates who went down this route. Unfortunately, many students achieved no marks on this part due to a wide variety of errors at the beginning. Calculating cos (360) on the calculator in radians instead of degrees was seen. Some tried to split the summation incorrectly into ∑ ቀ 3 4ቁ 𝑛𝑛 ∞ 2 x ∑ cos 180𝑛𝑛 ∞ 1 and find the sum to infinity of the trigonometric terms separately. A few wrote ( 3 4)2cos (360) but did not evaluate the cos(360). There were some very good solutions but in many responses there was often a lack of clear structure to their proof.

Common mistakes

• cos(180n)° = (−1)ⁿ so r = −3/4
• first term is (3/4)² = 9/16 because the sum starts at n = 2
• calculator in radians gives wrong values
• do not split into two separate sums

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question on modelling the speed of a car using arithmetic and geometric sequences proved popular. Many candidates answered both parts well and gained full marks. A small number of candidates used the sum formula in both (a) and (b) and as a result scored no marks. Additionally other candidates used their answer from (a) as one of the terms in (b), leading to an incorrect value of r. The second method mark was made available to these candidates.

Common mistakes

• 115 = 28r⁵ (use 1st and 6th gear, not the answer to (a)), r = 1.3265
• 5th gear 28r⁴ = 86.7 km/h
• premature rounding of r

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

A number of the responses were completely blank suggesting perhaps that candidates had run out of time to complete this question. In part (a), very high proportion of the candidates had some idea what the layout of the proof looked like yet only around 50% were able to complete it accurately and rigorously. There were many solutions containing the final the term of Sn as arn rather than arn - 1. Candidates who had the correct initial sequence regularly went on to complete the proof correctly. Only a small number prematurely factorised. Some were trying to apply the method of proving the sum of an arithmetic series instead, hence losing the majority of the marks.

Common mistakes

• last term is ar^(n−1), not arⁿ
• multiply by r and subtract
• both sides seen unfactorised
• do not use the arithmetic series method

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

A number of the responses were completely blank suggesting perhaps that candidates had run out In part (b), the first method mark was gained by the majority of students. Some didn’t divide by “a” or wrongly cancelled r10 and r5 in their equation, so lost the first accuracy mark. However, many of the students who gained the first two marks continued to gain the last two marks by dealing with the quadratic and achieving the correct solution for “r”. A few candidates used (n - 1) in the formula rather than n and a number of candidates got confused with n and r, replacing r = 10 and 5 in the denominator of each fraction. Only around 50% realised that they had ended up with a quadratic equation in r5 and candidates who did not cancel the "a" out early often ended up with an overcomplicated equation which they could not solve. Very few did not eliminate r = 1 if they found it, although a few solved the quadratic and then forgot to cube root the answer and so had r = 3.

Common mistakes

• cancel a and (1 − r): 1 − r¹⁰ = 4(1 − r⁵)
• quadratic in r⁵
• reject r = 1
• r = fifth root of 3 (not r = 3)
• S₁₀ = 4S₅, not S₅ = 4S₁₀

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Overall very good attempts to this question were spoiled by a lack in accuracy . Part (a) was routine for most and candidates were thorough in showing their method. The most frequent and concise method involved calculations of the form 2 24 6 1.05 6 1.05 . Occasionally the work was overcomplicated by converting to seconds though the end result was generally satisfactory.

Common mistakes

• 24 + 6 × 1.05 + 6 × 1.05² = 36.915 min = 36 min 55 s

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Overall very good attempts to this question were spoiled by a lack in accuracy . Part (b) was found to be challenging with some candidates choosing not to answer. The most straightforward way was to use some values and spot the patterns. This is a skill taught at GCSE yet seemed forgotten to many. Time taken Using the pattern on the left the time taken for the r th kilometre would be 4 6 1.05r

Common mistakes

• write out the 5th, 6th, 7th km times to see the pattern 6 × 1.05^(r − 4)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Overall very good attempts to this question were spoiled by a lack in accuracy . 5th kilometre 6 1.05 1 6th kilometre 6 1.05 2 7th kilometre 6 1.05 3 Although most candidates knew that for part (c) they needed to sum a sequence of terms, errors were frequent and many. A sizeable minority lacked any appreciation of the model and resorted to the calculation 20 6 1.05 1 1.05 1 . The majority of errors however were mainly due to a failure to link up the correct values for a and n in the formula for a geometric series. The most common of these was to use the strategy 24 + 16 6 1.05 1 1.05 1 , forgetting that the geometric sequence in this case should start at 6.3 instead of 6. It is often a good idea in such a question to write out the first few terms to gain an appreciation of the model and how the sequence is formed. (See below) Km 1st 2nd 3rd 4th 5th 6th 7th 20th Time 6 6 6 6 6 1.05 2 6 1.05 3 6 1.05 16 6 1.05 From here it should be a fairly straightforward task to write down 24 + 16 6.3 1.05 1 1.05 1

Common mistakes

• 24 + 6.3(1.05¹⁶ − 1)/0.05
• first term 6.3 not 6 and n = 16
• wrong 6(1.05²⁰ − 1)/0.05 ignores the model
• 173 min 3 s

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Q08 was found challenging by many candidates, with Q08(i) more successfully answered than Q08(ii). The most successful candidates were the ones who listed the first few terms of the series. In other words, those candidates that wrote 4 5 6 3 4 5 1 1 1 log log log ... and gained a 20 20 20 ... 5 5 5 2 3 4 2 2 2 better understanding of the series that they were required to sum. There were, however, many attempts that made no creditable progress and a significant number of blank responses to either one or both parts of Q08. In Q08(i), many candidates identified that a sum to infinity was required in their 1 a r . solution and attempted to apply the formula Most used in this formula, 2 1 r 1 1 . r r although some used incorrect values such as or Disappointingly, a 4 2 r 4 significant number of candidates used from the expression given in the question a , and applied it to the formula even though the formula only works when 1 r 1 1. r There were a number of correct strategies that candidates could use. The 4 a 1 1 , most successful strategy was to apply to and many 20 , a r 1 r 2 2 candidates who used this strategy achieved the correct answer of 2.5. Other complete 3 r r 1 1 strategies included applying or 20 20 2 2 1 1 r r 3 r r 1 1 which led to (20 17.5) or (40 37.5) respectively. 20 20 2 2 0 0 r r Some candidates applied an incorrect overall strategy by subtracting an incorrect number of terms from their sum to infinity. These candidates gave incorrect answers such as (20 18.75), (40 38.75), (40 35) or (40 17.5). A few candidates obtained the correct answer of 2.5 by using their calculator. Only a few candidates produced a correct solution to Q08(ii). There were two main correct methods that were used by candidates in equal measure. One was to list the first few terms and the last few terms of the series and use the addition law of logarithms 3 4 50 log ... . to achieve Many candidates used cancelling to complete their 5 2 3 49 50 proof correctly by writing The other method was to use the log log (25) 2. 5 5 2 48 subtraction law of logarithms to give and list terms to (log ( 2) log ( 1)) n n 5 5 1 n (log 3 log 4 ...... log 50) (log 2 log 3 ...... log 49). give Again many 5 5 5 5 5 5 candidates used cancelling and completed their proof by writing A few candidates progressed by writing log (50) log (2) log (25) 2. 5 5 5 log (3 4 ....... 50) log (2 3 ....... 49) and used factorials to give 5 5 50! 50! log log 49! log log (25) 2. 5 5 5 5 2 2(49!) Some candidates incorrectly assumed an arithmetic series and attempted to find its sum 3 50 n a n 48, log , log n a l ( ) l [2 ( 1) ] a n d by using or with and 5 5 2 49 2 2 4 3 log log . d A few candidates incorrectly assumed a geometric series and 5 5 3 2 attempted to find its sum. Some candidates, who made no creditable progress, resorted to listing their terms as decimals, or attempted to add all 48 terms as decimals by writing 0.2519... 0.1787... 0.1386... ...... 0.0125... 2.

Common mistakes

• first term a = 20(½)⁴ = 1.25, r = ½, S∞ = 2.5
• listing the first terms helps
• subtracting the wrong number of terms from a sum from r = 0 or 1
• using r = 20 or ¼

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was an accessible question with part (ii) more successfully answered than part (i). 16 r (3 5 2 ) r In part (i), some students attempted to find by applying the strategy 1 r 16 16 16 16 r r (3 16) 5 2 , r (3 5 ) 2 . r while other students applied the strategy Most 1 1 r r 1 1 r r 5, 5 a d students applied the sum formula for an arithmetic series with to 16 16 8, 5 a d (3 16) 5 r (3 16) (2(5) 15(5)) to give while some applied to 2 1 r 16 n 1 ( to give 16 (2(8) 15(5)). (3 5 ) r 1) r n n A few students used to help them to 2 2 1 r 1 r 16 5 . r evaluate Students were generally less successful with using the formula for the sum 1 r 16 r r 2 . of a geometric series to find Common errors included assuming the first term, a, was 1 1 or writing down the value 131070 with no method (perhaps by working back from the given answer). A significant number of students listed all 16 terms and added them to achieve the given result. Those students who did not list all the terms, but just wrote e.g. 10 17 26 39 60 ... 65619 131798, did not receive any credit. Some students, who attempted to find the first, second and third differences between the terms of the series 10, 17, 26, 39, 60,..., usually made no creditable progress. Part (ii) was very well attempted, even by those who failed to gain any credit in part (i). Almost all students wrote down the next two terms u correctly. Many deduced that they , u 2 3 100 were working with an alternating sequence and found the correct exact value for . u r 1 r Some students attempted to round their answer, usually to 108.3. A few students attempted 100 to find by using the formula for the sum of an arithmetic series or the formula for the u r 1 r sum of a geometric series.

Common mistakes

• split into Σ(3 + 5r) + Σ2^r
• AP sum 8(8 + 83) = 728
• GP sum with a = 2 (not 1), r = 2: 2(2^16 − 1) = 131070
• listing terms needs all 16 shown