P4.4 Arithmetic series

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

Examiner's comment

This question tested arithmetic sequences, followed by finding the sum for a given number of terms. Part (a) was usually answered well. There was a variety of ways which could be used to set up a valid equation or equations from the information given. The most common method was equating the common difference between consecutive terms resulting in a single linear equation that was easily solved for k. Some responses used slightly less efficient methods by using the nth term formula which created two linear equations in k and the common difference, d, which they then solved simultaneously. A few responses used the sum formula in addition to an nth term formula which proved quite time consuming and often incurred errors when solving. A minority of responses mistakenly treated the sequence as geometric and gained no marks. Occasionally learners deduced the value of k with little or no working by spotting that the second term must be 4k in order that the sequence was arithmetic (2k, 4k, 6k) and then simply divided 10 by 4.

Common mistakes

• equal common differences 10 − 6k = 2k − 10 (or 10 is the mean of 6k and 2k) gives k = 5/2
• treating it as geometric scores 0

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

Examiner's comment

This question tested arithmetic sequences, followed by finding the sum for a given number of terms. Part (b) was also well attempted, although more errors were noted here. The responses seen usually found the values for the first term and common difference by substituting their value of k back into the original sequence. A few mistakenly used d = 5 as they did not acknowledge that the sequence was decreasing. Some simply used their value of k in place of d in the sum formula which gained no marks at all. The responses that had set up two linear equations in part (a) had often already evaluated d in part (a) when they were solving their simultaneous equations, but it was only when it was seen or used in part (b) that they gained credit. While full marks were often earned in this part, errors in applying a correctly quoted formula sometimes resulted in marks not being gained. For example, (50 – 1) – 5 instead of (50 – 1)(–5) commonly resulted in 44 instead of –245 which lost the final mark. A few responses adopted an alternative approach of using 5 20 n - + ∑ and either used their calculators or standard sum formulae from Further Maths to evaluate it. Those who had treated the sequence as geometric in part (a) often continued to do so in part (b) gaining no marks, although a small number of responses did use their surd value of k to correctly find follow through values for the first term and common difference of an arithmetic sequence. A very small minority opted for the primitive method of writing out all 50 terms and adding them together individually. They were sometimes successful, but this was an extremely time- consuming method and would have compromised the time available for other questions on the paper. 7

Common mistakes

• a = 15, d = −5 (the sequence decreases
• d = 5 and using k as d are common errors)
• S₅₀ = 25(30 + 49 × (−5)) = −5375
• bracket slips such as (50 − 1) − 5

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

Examiner's comment

This question provided some challenge and discrimination for the end of the paper. Most learners attempted this and made some progress. Part (a) was a “show that” question and learners need to remember therefore to show all steps in their working clearly and present a solution without incorrect working. Most were able to identify that they needed to use the fact that the sum of the probabilities equalled 1. A popular starting point was to write down 1= 2 4 6 2 ... n k k k k + + + + and some were then able to recognise that an arithmetic series was involved and use a suitable formula to find the sum. Some learners didn’t use the correct number of terms with 2n or 4 (from the 4 terms 2, 4, 6 and 2n) being common errors. Full and concisely presented solutions were rare.

Common mistakes

• sum of probabilities = 1: (2 + 4 + ... + 2n)/k = 1 with n terms (not 2n or 4)
• arithmetic series sum n(n + 1)
• every step shown for the given answer

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

Examiner's comment

Part (a) was very accessible and very well answered. The vast majority of candidates deduced the value of the common difference and then substituted this into the standard term formula for an arithmetic sequence. Those candidates that were unfamiliar with this formula were able to produce a correct calculation, or even just list the payments for all 12 months, which was acceptable. Occasionally the formula was simplified to 410 - 10n before substituting in n = 12.

Common mistakes

• d = −10
• u₁₂ = 400 + 11(−10) = 290 with brackets correct
• listing all 12 terms is acceptable

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

Examiner's comment

Part (b) was usually answered well, and candidates were able to substitute the given values into the sum formula correctly, expand brackets and obtain the required quadratic equation in N. The fact that the common difference was negative caused some candidates algebraic difficulties. Some candidates did not realise they needed to set their expression equal to 8100 but soon realised this and added it in later in their working but a few persisted with e.g. = 0 rather than = 8100. A few had bracketing and/or arithmetic errors and some missed out the ‘= 0’ in the given answer therefore losing the final accuracy mark. A small number of candidates attempted to use the term formula in this part. Another rarer approach involved using a series summation approach. These attempts were mostly successful as it required further maths knowledge and were therefore performed by most proficient mathematicians.

Common mistakes

• set S_N = ½N(800 + (N − 1)(−10)) equal to 8100 (not 0)
• negative d causes sign slips
• expand all brackets and include "= 0" in the given answer

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

Examiner's comment

Candidates found part (b) more challenging, with many failing to link areas of mathematics, in this case trigonometry and series. Even where a candidate successfully found the sum, 9cos 36sin S x x = + , many did not relate it to part (a). Those who did recognise the form were usually able to get 4.5 " " R , usually 4.5 2 17 for part (i). Some seemed to misread the question and stated the minimum value, so lost the accuracy mark. A significant number of candidates found a correct expression in harmonic form but lost both the M1 and A1 as they kept a correct 9 17 embedded in their expression.

Common mistakes

• common difference sin x: S₉ = 9cos x + 36sin x = 4.5(2cos x + 8sin x), so max = 4.5R = 9√17 (not left embedded in an expression)
• occurs when x = α ≈ 1.33
• many did not link to (a)
• giving the minimum

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

Examiner's comment

This question on arithmetic sequences and series proved to be very discriminating. The proof of the sum of an arithmetic sequence is standard, but was not answered well by many candidates. Most candidates however were able to then made progress with the rest of the question, having been given the required formula. Good candidates often did well in this question however, with 17% of the cohort achieving full marks. In part(i) those who had learned the standard proof for the sum of the first n terms of an arithmetic series often gained full marks. It was however, unusual to see a well set out, fully correct proof. Candidates often failed to write a correct expression for Sn at the start of their proof, or sometimes having an incorrect first or last term or extra terms (most often a+nd) within their sum. Other errors included not displaying a sufficient number of terms, not writing S= or Sn =, and using commas instead of + sign between the terms. In part(ii) most realised that they could make headway even if they hadn’t attempted part (i). In (ii) (a) many of the candidates used a=20, d = -0.8 and Sn = 64 in the sum formula to score the M mark. Poor notation such as (n-1)-0.8 was frequently seen, but candidates usually recovered from this. Following a correct equation most were able to proceed correctly to the given quadratic for the accuracy mark. A common error was where candidates used +0.8 as the common difference rather than -0.8 Part (ii) (b) was generally answered without issues. Almost all candidates managed to get B1 for getting the two values of n, either by factorising or by using a calculator. Part (ii)(c) was not answered very well with most candidates identifying 10 as the valid solution but not adequately justifying this choice. The reasons provided were often too vague or not related to why 10 was better than 16. More tended to score the mark by explaining that at 16 weeks he would be saving a negative amount rather than using the fact that 10 was the smaller number or that he would have had enough by 10 weeks so would not continue saving until 16 weeks.

Common mistakes

• write Sn = a + (a + d) + ... + (a + (n − 1)d) with + signs, reverse and add to get 2Sn = n(2a + (n − 1)d)
• wrong extra terms such as a + nd
• too few terms shown

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

Examiner's comment

This question on arithmetic sequences and series proved to be very discriminating. The proof of the sum of an arithmetic sequence is standard, but was not answered well by many candidates. Most candidates however were able to then made progress with the rest of the question, having been given the required formula. Good candidates often did well in this question however, with 17% of the cohort achieving full marks. In part(i) those who had learned the standard proof for the sum of the first n terms of an arithmetic series often gained full marks. It was however, unusual to see a well set out, fully correct proof. Candidates often failed to write a correct expression for Sn at the start of their proof, or sometimes having an incorrect first or last term or extra terms (most often a+nd) within their sum. Other errors included not displaying a sufficient number of terms, not writing S= or Sn =, and using commas instead of + sign between the terms. In part(ii) most realised that they could make headway even if they hadn’t attempted part (i). In (ii) (a) many of the candidates used a=20, d = -0.8 and Sn = 64 in the sum formula to score the M mark. Poor notation such as (n-1)-0.8 was frequently seen, but candidates usually recovered from this. Following a correct equation most were able to proceed correctly to the given quadratic for the accuracy mark. A common error was where candidates used +0.8 as the common difference rather than -0.8 Part (ii) (b) was generally answered without issues. Almost all candidates managed to get B1 for getting the two values of n, either by factorising or by using a calculator. Part (ii)(c) was not answered very well with most candidates identifying 10 as the valid solution but not adequately justifying this choice. The reasons provided were often too vague or not related to why 10 was better than 16. More tended to score the mark by explaining that at 16 weeks he would be saving a negative amount rather than using the fact that 10 was the smaller number or that he would have had enough by 10 weeks so would not continue saving until 16 weeks.

Common mistakes

• a = 10, d = −0.8: 64 = n/2(20 − 0.8(n − 1)) then expand and simplify to the given quadratic

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

Examiner's comment

As the first question of the paper this did its job in settling nervous students: a straightforward question on arithmetic series which was answered correctly by the vast majority of students. Where mistakes were made the most common error in part (a) was to use the 21st term as 21, rather than 24, or students used an incorrect formula for the nth term, using un = a + nd and not un = a + (n -1)d. A few students incorrectly treated 24 as the sum of the first 21 terms in their attempt to find d.

Common mistakes

• uₙ = a + (n − 1)d, not a + nd
• the 21st term is 24, not 21
• 24 is a term, not the sum of 21 terms

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

Examiner's comment

As the first question of the paper this did its job in settling nervous students: a straightforward question on arithmetic series which was answered correctly by the vast majority of students. For part (b) most students used the correct formula and therefore gained the method mark, however some lost the accuracy mark from the incorrect common difference, and a very small minority of students used 𝑆𝑆500 = 𝑛𝑛 2(a + (n - 1)d) rather than 𝑆𝑆500 = 𝑛𝑛 2(2a + (n - 1)d).

Common mistakes

• Sₙ = ½n{2a + (n − 1)d} with a = 16, n = 500
• using a instead of 2a
• wrong d from (a) loses the A mark

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. In (a) the most common errors were to divide ( ) 115 28 - by 6 in finding d. In part (b) errors were mainly due to a lack of accuracy and prematurely rounding their value for r.

Common mistakes

• 115 = 28 + 5d (not 6d, not (115 − 28)/6), d = 17.4
• 3rd gear is 28 + 2d = 62.8 km/h
• using the sum formula scores 0

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

Examiner's comment

students who had faltered on (b) still making a start on part (c). Many identified the 4 angles as terms of an arithmetic series and some were able to reach the condition 2a + 3d = 180 (or equivalent) but many could not see how to complete the problem. Some used the information that a > 50 to reach d < 80 3 but were unable to make the connection that this meant that only the two cases d = 10 and d = 20 were required and so the probability could simply be written down.

Common mistakes

• angles a, a + d, a + 2d, a + 3d sum to 360 so 2a + 3d = 180
• a > 50 only for d = 10 or 20
• P(D = 10 or 20) = 3k/20 = 90/137 exact

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

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

Examiner's comment

P(H = 5) rather than P(F = 5). Part (d) proved more challenging; those who carefully started to list the probabilities were often able to add these manually (few identified the arithmetic series as intended) and those who realised that the sum of the probabilities was equal to 1 usually went on to score all the marks and the mark in part (e). However, many thought that P(F = 10) = 1 or even that P(F = 5) had to equal their answer to part (c) and made little progress in (d) but could often collect the mark in (e). Part (f) was not answered well; the question clearly steered students to comment on the

Common mistakes

• list P(F = n) for n = 1 to 10 and use sum of probabilities = 1 (arithmetic series 5(0.02 + 9α) = 1) giving α = 0.02
• errors: assuming P(F = 10) = 1 or P(F = 5) equals answer to (c)