Edexcel A-level Maths · Topic 4
Practice questions covering sequences and series from this specification. Every question below is the real Pearson question as it is printed on the paper, with its mark scheme.
6 questions on Sequences and series, each with its mark scheme and worked method. Reveal the answer when you have had a go.
1.
(3)
(Total for Question 1 is 3 marks)
2.
(4)
(Total for Question 2 is 4 marks)
3.
(3)
(Total for Question 3 is 3 marks)
4.
(5)
(Total for Question 4 is 5 marks)
5.
(5)
(Total for Question 5 is 5 marks)
6.
(3)
(Total for Question 6 is 3 marks)
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Printable question pack — 4 Sequences and seriesAll 60 questions across 6 specification points — 54 more than the 6 above — with the full mark scheme. Guest accessible · browser print.33 questions shown as printed · newest series first
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Edexcel 9MA0 · June 2024 · Paper 1 · Q2 · 4 marks
4 marks
Examiner report
This was another question that was familiar to candidates, providing excellent access, with plenty of candidates scoring full marks and many more scoring 3 marks out of the available 4. In part (a), the presentation of the binomial structure was generally very good. Inconsistency with brackets or lack of attention to detail (often not simplifying correctly) were the main reasons why candidates who started correctly dropped accuracy marks for their final answer. A small minority attempted to use the version of the binomial expansion where the power is required to be a positive integer and made no progress, while some candidates ignored the coefficient of x, simplifying the problem and thus scored no marks. Candidates were able to recover these errors in subsequent work, but it was not common for this to happen. The most common error was using 9 instead of –9 which was allowed to score the method mark. A small number of candidates employed a Maclaurin expansion which was acceptable but should not be encouraged. Equally, although it was possible to use direct expansion to solve the problem these attempts by candidates were usually unsuccessful, particularly for those that tried to take out a factor of 9. Occasionally, candidates multiplied their correct simplified expansion by a constant such as 16 at the end, this could still achieve full marks as examiners could apply ISW. Finally, some candidates attempted to use notation of the form 1 2 2 but the majority using this did not evaluate it to a suitable form to be able to score the method mark. Part (b) proved to be more challenging for candidates. Candidates who stated an acceptable range of validity were usually able to compare this to 2 9 − and correctly conclude that the expansion was not valid. However, there were also incomplete attempts where the range of validity was found but without a correct comparison or conclusion. Candidates often struggled to correctly identify the range of validity giving answers such as 1 9 x − . Quite a few responses were blank, while those who attempted it fell evenly between those who gave vague or incorrect responses, and those who recognised what the question was asking. Amongst those candidates, the mark was often missed due to a misunderstanding of the nature of the modulus function, leading to incorrect statements such as x > 1 9 or 1 2 9 9 − . Many attempts to evaluate the expansion by substituting 2 9 x = − into 2 3 9 81 729 1 2 8 16 x x x − − − to give 2, then stating 2 3 were seen and these did not score the mark.
This question concerns the binomial expansion of (1 - 9x)^(1/2).
Edexcel 9MA0 · June 2024 · Paper 1 · Q9 · 6 marks
6 marks
Examiner report
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. 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.
The first three terms of a geometric sequence are 3^(4k-5), 9^(7-2k), 3^(2(k-1)), where k is a constant.
Edexcel 9MA0 · June 2024 · Paper 2 · Q2 · 5 marks
5 marks
Examiner report
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. 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. In part (c), most candidates were able to solve the given quadratic equation and achieve the method mark. Overall, this was done with a calculator. Most chose the correct answer but quite a few left both answers or chose N = 45, appearing not to take into account the context of the question. They needed to realise that, when the loan was fully repaid after 36 months, the borrower would not continue to make loan repayments, so the second answer of 45 months needed to be rejected. This took significant effort for some - rechecking the sum for 36 and 45 terms and then finding the value of the 36th and 45th term and then rejecting 45 when the 45th term was negative - when thinking about the context would have been easier.
Jamie takes out an interest-free loan of £8100. Jamie makes a payment every month to pay back the loan. Jamie repays £400 in month 1, £390 in month 2, £380 in month 3, and so on, so that the amounts repaid each month form an arithmetic sequence.
Edexcel 9MA0 · June 2024 · Paper 2 · Q4 · 5 marks
5 marks
Examiner report
Generally, candidates found this question accessible, with many scoring full marks or nearly full marks. In part (a) most candidates used the given recurrence relation along with u1 = 6 to find u2 = 6k – 5 and then u3 = 6k2 – 5k – 5, and then set u3 = – 1 and rearranged to obtain the given answer 6k2 – 5k – 4 = 0. Algebraic errors were few, but some candidates forgot to include the ‘= 0’ in their final line and lost the second mark in this part. There were occasional attempts at using arithmetic or geometric series which gained no credit. In part (b)(i) most candidates were able to solve the quadratic equation, but not all identified that k = 4 3 was the only correct answer, or subsequently used this value, so lost the mark in this part of the question. Some candidates spent more time than they needed using the quadratic formula, where a factorisation of the quadratic would have been more direct. Use of the calculator equation solver was also permitted here. Candidates must learn to check their answers against any limitations given in the question. When errors did occur, they were most likely to appear in part (b)(ii), although most still answered it well. Most candidates identified the requirement of the question to add the first three terms of the sequence together, and correctly used k = 4 3 to calculate u2 =6 ( 4 3 ) -5 before finding 6 + 3 + (−1) = 8. Others first simplified the sum to obtain 6k or 6k2 + k − 4 before substituting the value of k. A common incorrect approach was to use u3 = 6k2 – 5k – 4 in the sum. Some also used k = 4 3 and k = 1 2 − to calculate the sum, without identifying which was the required answer. Alternatively, a good number of candidates attempted, correctly and incorrectly, to work out the values of u1 and u3 even though they were given in the question. It was also noticeable that a number of candidates used the quadratic expression from part (a) in order to work out values in (b)(ii), usually assuming u3 = k2 – 5k – 4. They, therefore, did not clearly understand what this equation represented. A few candidates incorrectly attempted to use the recurrence relation to find things like ( 4 3 )(1) – 5, ( 4 3 )(2) – 5 and ( 4 3 )(3) – 5, using these expressions to find their sum, and there were again occasional attempts using arithmetic or geometric series which also gained no credit. There were also a few attempts at integration. Overall, this was a question that allowed candidates to gain a good number of marks with confidence.
A sequence u_1, u_2, u_3, ... is defined by u_(n+1) = ku_n - 5, with u_1 = 6, where k is a positive constant. Given that u_3 = -1:
Edexcel 9MA0 · June 2023 · Paper 1 · Q9 · 7 marks
7 marks
Examiner report
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. 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.
The first three terms of a geometric sequence are 3k+4, 12-3k and k+16, where k is a constant.
Edexcel 9MA0 · June 2023 · Paper 2 · Q2 · 6 marks
6 marks
Examiner report
For the most part candidates answered this question well. This was dependent on whether they correctly understood the iteration formula. In part (a) (i), although many scored this first mark, there was a common misunderstanding of how a recurrence relationship works with confusion as to the meaning of n. Rather than term position, it was often confused with 𝑢𝑛 so that the incorrect calculation ( ) 35 2 35 35 7cos 5 1 2 u = + − − was often seen, giving the correct answer but obviously not scoring the B mark. Also not uncommonly seen was n replaced by 1 n u + , with candidates stating ( ) 40 2 40 40 7cos 5 1 2 u = + − − . The errors stated above also affected the performance of some candidates in part (ii). Furthermore, a minority of candidates had difficulty with accuracy in the sign of the final term, struggling with the odd/even power of −1 . For example, when calculating 3 u , ( ) 2 3 2 40 7cos 5 1 40 7 5 38 2 u = + − − = − + = , was not uncommon. Occasionally, candidates were seen to be working with degrees rather than radians, also leading to loss of the accuracy mark. In part (b) (i) most candidates understood the order 4, periodic nature of the recurrence relationship and that 5 1 4 3, k u u u k Z + − = = . Occasionally, candidates unnecessarily calculated 5 u , using 4 u in the recurrence relationship, which sometimes led to the mark being lost either if 4 u had an incorrect value or if an error was made in the calculation. In part (ii) many candidates recognised how to construct the arithmetic calculation to find the required sum. The most common of these attempts were: ( ) 1 2 3 4 1 6 u u u u u + + + + and ( ) 2 3 4 1 6 7 u u u u + + + Credit was given to candidates who used such ideas with incorrect values from earlier in the question for 3 u and 4 u , although those who had a wrong answer in (a)(i) who used their incorrect value for 2 u (rather than the given 2 40 u = ), lost both marks. It is important that centres make candidates aware that they must use the ‘show that’ answer in subsequent calculations no matter what they achieve in their working. Occasionally, candidates used ( ) 25 1 2 3 4 1 6.25 r r u u u u u = = + + + but these usually did not add 1 to give a fully correct method. Some also incorrectly tried summing the series as though the terms formed an arithmetic progression. Some candidates incorrectly multiplied (u1 + u2 + u3 + u4) by 4 instead of by 6. A minority forgot to add 35 to their total. A very small minority of candidates generated all the terms of the whole sequence and added them together.
A sequence u_1, u_2, u_3, ... is defined by u_1 = 35 and u_(n+1) = u_n + 7 cos(n*pi/2) - 5(-1)^n.
Edexcel 9MA0 · June 2023 · Paper 2 · Q8 · 6 marks
6 marks
Examiner report
Part (a) of this question was generally answered very well with candidates regularly scoring full marks. Candidates seemed to be well practiced at converting to harmonic form and working in radians. The most successful candidates wrote out the full expansion of ( ) cos R − rather than trying to take a short cut. The common issues included giving a decimal solution for R or rounding the value of α to a lower degree of accuracy than was required in the question. There were a small number of candidates who worked with 2 tan 8 = and so lost the method mark and the accuracy mark. 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. Candidates were slightly more successful in part (ii). Most candidates who stated a correct value for the maximum, also got the value of x correct, more often than not giving it to the same accuracy as in part (a). A few benefitted from this being a follow through mark. Some candidates tried to find a subsequent value for and were often out by or 2, or attempted to rearrange 1.326 x − = . A few responses were seen where S was differentiated and set to zero to find a maximum. These candidates often lost the accuracy mark as they found an answer of 37.1. A lot of incorrect responses gave a maximum of 36 when sin 1 x = or 45 from finding 36 + 9.
The first three terms of an arithmetic sequence are cos x, cos x + sin x, and cos x + 2 sin x, where x is not an integer multiple of pi. Let S_9 be the sum of the first 9 terms as x varies.
Edexcel 9MA0 · June 2023 · Paper 2 · Q13 · 13 marks
13 marks
Examiner report
Most candidates attempted at least part (a) of this question, but some found the work more demanding in (b) and (c) and made little progress in either part. Despite this, there were many fully correct or nearly fully correct attempts at the whole question. In part (a), most candidates successfully factorised out 2 3−and attempted the correct resulting expansion of 2 1 3 x − + . There were some arithmetic slips, but generally candidates often reached the correct answer. A common error, however, was failing to factorise properly, reaching ( ) 2 2 3 1 . x − − + There were a number of attempts at direct expansion. Some of these used non-standard binomial coefficients involving negative integers. To achieve credit for using these, candidates needed to demonstrate the meaning of their non-standard notation. Very occasionally candidates multiplied their correct simplified expansion by 9 or 27 at the end. This still received full marks as examiners could ignore subsequent working following seeing a correct simplified answer. In part (b), many candidates correctly multiplied their expansion by 6x and then integrated using the limits of 0.2 and 0.4 reaching 0.03304, gaining full marks. A common error, however, was not giving the final answer to the required accuracy of four significant figures, thus losing the final accuracy mark. Those candidates who had expanded part (a) to more than three terms also lost this final accuracy mark as it led to a different approximation. A more complicated and unnecessary method involving integration by parts was used by a small minority of candidates. Candidates also often tried inappropriate approaches like dividing 6x by their expansion followed by a logarithmic integral. A minority of candidates, despite the warning at the start of the question, clearly used calculator technology to obtain 0.032865, rather than the answer of 0.03304 required. Others made inappropriate attempts at the trapezium rule, thus gaining no credit as algebraic integration was required. In part (c), the commonest methods of integration adopted were substitution, using either u = x + 3 or u = (x + 3)2 , integration by parts, integration after applying partial fractions and integration using the reverse chain rule after writing the integrand in appropriate form. Candidates had varying success with all of these approaches, gaining a variety of marks. Integration by parts seemed to lead to most arithmetical mistakes. Errors resulted from using parts ‘the wrong way round’ and others integrated the ( ) 2 3 x − + term incorrectly. When integrating by substitution some students incorrectly thought that 2 18 u − would integrate to a natural logarithm. There was some confusion with the limits; some candidates did not change the limits to “u” values and other who did evaluate new limits for their substitution switched back to an expression in x before substituting their changed limits. When using partial fractions some students struggled with the initial format for the partial fractions and it was not uncommon to see partial fractions of the form 3 3 A B x x + + + . There was a number of candidates who either did not attempt this part of the question at all or who lacked a correct strategy. There were responses where students attempted to integrate numerator and denominator separately, e.g. 6x to 3x2 and ( ) 2 1 3 x + to a variation of ( ) 2 ln 3 x + , and proceeded to multiply both results. A number of candidates were awarded just the first mark as their overall problem solving strategy was correct despite the fact they were unable to fully implement it. There were also many candidates who could carry out the work in this part completely successfully.
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
Edexcel 9MA0 · June 2022 · Paper 1 · Q13 · 7 marks
7 marks
Examiner report
(Mean mark 3.8 out of 7) 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.
In an arithmetic series, the first term is a and the common difference is d.
Edexcel 9MA0 · June 2022 · Paper 2 · Q3 · 4 marks
4 marks
Examiner report
This question proved to be successful for many candidates and many scored full marks here. In part (a)(i), the vast majority were able to successfully write down the sequence of terms and hence demonstrate that the sequence was periodic. An algebraic proof was not required here, and one was only rarely seen. Perhaps surprisingly having identified the pattern in the sequence in (a)(i), not all were able to identify the order of the sequence as ‘2’ in (a)(ii). This perhaps highlights a lack of awareness of what is meant by “order” of a sequence. Common incorrect answers here were: ‘first order’; ‘order 3, 5’; ‘±2’or just a re-writing of their sequence from part (a)(i). In part b), candidates were most successful when they considered the nature of the sequence rather than trying to resort to a formula. A fairly common first step was to divide 85 by 2 to give 42 with remainder 1. Many students then correctly broke down the sum into either 42 × 3 + 43 × 5 or equivalently (42 × 8) + 3 . Errors occurred when candidates struggled with the odd number of terms and believed they should calculate 42.5 × 5 + 42.5 × 3 resulting in an answer of 340, or when candidates mistakenly thought that the 85th term was ‘5’ rather than ‘3’ resulting in an answer of 341. Other errors included summing only 84 terms (42 × 8 = 336) or attempts to ‘chunk’ incorrectly into blocks of e.g. size 5: in such attempts the sum of the first five terms was obtained (3 + 5 + 3 + 5 + 3 = 19) followed by deduction that 85 5 = 17 leading to an incorrect answer of 19 × 17 = 323. A sizeable number of incorrect attempts involving use of the sum of arithmetic or geometric series were also commonly seen. Perhaps most surprising was the significant minority who were able to set up a correct approach and write down a correct method but who then made arithmetical errors to write down an incorrect total which cost them the final A mark.
A sequence a_1, a_2, a_3, ... is defined by a_1 = 3 and a_(n+1) = 8 - a_n.
Edexcel 9MA0 · June 2022 · Paper 2 · Q7 · 5 marks
5 marks
Examiner report
This question on binomial expansion was well attempted and generally well answered by most candidates. In part (a), almost all candidates achieved at least 1 mark, either by extracting the factor of 41/2 correctly, or for having a correct binomial coefficient multiplied by the corresponding power of x on their 3rd or 4th term. A few candidates omitted the −9/4 and wrote ቀ1 2ቁቀ1 2−1ቁ 2! 𝑥𝑥2 as their 3rd term, which was not awarded the method mark. Some candidates used 9/4 erroneously, instead of −9/4 in their expansion, and this could still gain the method mark but meant the two accuracy marks were lost. Candidates bracketing was sometimes incorrect but often recovered by later work. A common response seen was a correct expansion of ቀ1 − 9𝑥𝑥 4 ቁ 1/2 but failure to multiply this by the “2” they had factorised outside the bracket at the beginning. A minority of candidates attempted a direct expansion method, and many of these achieved the first mark only for a correct first term, as they often did not find the correct binomial coefficients. Unfortunately, several candidates made number of slips, or omitted an exponent, in their final line and did not write the correct answer following correct working and so lost the final accuracy mark. In part (b) very few could articulate correctly why this was an overestimate. Many left this part out or just wrote overestimate or underestimate with no explanation. Some were very inventive in their reasons. Poor wording/clarity on subtracting terms, such as ‘decreasing trend’, or ‘terms decreased’ but not specifically stated to be subtracted from the leading 2. Many tried substituting in values to show it was smaller which was not sufficient. The majority did not relate the reasoning back to the binomial expansion directly. Some went on to evaluate the binomial using x = 1/9. Some said the terms were getting smaller rather than pointing out that they were negative. Some, however, gave a very clear reasoned argument. Several students answered part (b) with an explanation about the validity of the expansion using |x|<1. Some students thought the overestimation was due to the fact that the expansion had only used 4 terms and that the approximation would be better if more terms were used, with no reference to the fact that they should be subtracted from 2. A common mistake was stating that 𝒙𝒙< ቂ 𝟒𝟒 𝟗𝟗ቃ which in their opinion meant that the approximation was over/underestimation due to the fact that 𝟒𝟒 𝟗𝟗< 𝟏𝟏 𝟗𝟗 or the other way round.
Consider sqrt(4/9 - x). A student uses its binomial expansion with x = 1/9 to find an approximation for sqrt(3).
Edexcel 9MA0 · June 2022 · Paper 2 · Q15 · 10 marks
10 marks
Examiner report
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). Part (a) was attempted successfully by most candidates, starting usually with 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐= 6𝑡𝑡𝑎𝑎𝑎𝑎𝑎𝑎 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠. Candidates then successfully cross multiplied and used the appropriate identity to reach the required result. Another starting point regularly seen was 6𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡= 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቀ 5+2𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 12𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐ቁ 2 . Errors in the proof were rare if candidates started from a correct statement. Occasionally the “ = 0” was carelessly missing. In part (b) most candidates correctly solved the equation either by factorising or using calculator technology, but many reached the acute angle 𝜋𝜋 6 rather than the required obtuse angle of 5𝜋𝜋 6 . Others found a number of solutions but failed to identify 5𝜋𝜋 6 as the required answer. A few candidates worked in degrees and again lost the accuracy mark. Those who had obtained an angle of 𝜋𝜋 6 in (b) then usually carried this error into (c) and were in a position only to earn the method marks. Candidates using 𝜋𝜋 6 in (c) often would reach 9൫1 + √3൯, but did not seem to worry that this did not match the required answer 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 common ratio. Some candidates unfortunately put effort into finding and rearranging (often incorrectly) algebraic expressions for the sum to infinity involving 𝜃𝜃, without substituting an angle in, gaining no credit, and others used 𝑎𝑎 1+𝑟𝑟 incorrectly as the formula for the sum to infinity. The most common mistake in (c) was failure to show the steps where the denominator was rationalised. Many candidates clearly typed an expression like −6√3 1−൬−√3 3 ൰ straight into their calculators to obtain 9 −9√3 losing the last two marks in the question.
In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable. The first three terms of a geometric series are 12cos(theta), 5 + 2sin(theta), and 6tan(theta).
Edexcel 9MA0 · October 2021 · Paper 1 · Q3 · 6 marks
6 marks
Examiner report
This proved to be a little more demanding than the first two questions but over 40% of candidates still scored all 6 marks. Errors seen by candidates are summarised below. In part (a) several candidates did not simplify 𝑢𝑢2 sufficiently, giving 𝑢𝑢2 =𝑘𝑘−24/2 which then lead to a more complicated expression for 𝑢𝑢3 which they had difficulty simplifying. Some candidates did not use the given expression, but rather used 𝑢𝑢1+𝑢𝑢2+𝑢𝑢3 = 0. In part (b) candidates usually found both roots of the given quadratic either by factorisation or by use of the quadratic formula, with a small number by completing the square. A significant minority failed to gain the A mark here as they did not state that 6 was an integer or that 40/3 was not an integer, with some excluding 40/3 on the grounds that it was irrational. Part (c) was invariably answered correctly by those who had solved the quadratic in part (b). although there were some candidates who gave a choice of answers.
The sequence u_1,u_2,u_3,... is defined by u_(n+1)=k-24/u_n, with u_1=2, where k is an integer. Given that u_1+2u_2+u_3=0:
Edexcel 9MA0 · October 2021 · Paper 1 · Q5 · 6 marks
6 marks
Examiner report
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 20 000×1.082 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. 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. 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.
A company made a profit of GBP 20,000 in Year 1. A model predicts that yearly profit increases by 8% each year, so the yearly profits form a geometric sequence. Show all stages of working; solutions relying entirely on calculator technology are not acceptable.
Edexcel 9MA0 · October 2021 · Paper 1 · Q9 · 11 marks
11 marks
Examiner report
This question was answered with varying degrees of success. Around 28% of the candidates failed to score any marks here at all, which was very suprising. There were however, many well formed and accurate solutions by careful candidates. In part (a) most candidates managed to write down a correct identity and use an appropriate method to find values for B and C. To prove A =0 was generally done via equating coefficients of x2 or by setting x = 0 or x=1. A few candidates wrote down three equations for the coefficients and constants, but rarely was a correct set of solutions gained from this method. It was also not uncommon to see the RHS expanded either with or without the terms involving A in an attempt to prove that A = 0. Errors included having an incorrect identity from the start which resulted in no marks. In part (b)(i) it was rare to see fully correct expansions. Good candidates often lost the final mark due to an error in adding the required terms. A common error was forgetting to use their value for C and only adding the expansion for (1-2x)-1 rather than C(1-2x)-1. A few were unable to extract 2-1 correctly from (5x+2)-2 (usually having this as 2). Other errors seen involved an expansion of (1 + 2/5 x)-2 or else (1 + 5/2 x) 2 rather than (1 + 5/2 x) -2. Very few correct answers were seen for part (b)(ii) and this was very often not attempted. Those who did write a correct range sometimes also gave a second range which therefore did not gain the B mark here.
Let f(x)=(50x^2+38x+9)/((5x+2)^2(1-2x)), where x!=-2/5 and x!=1/2. Given that f(x) can be expressed as A/(5x+2)+B/(5x+2)^2+C/(1-2x), where A, B and C are constants:
Edexcel 9MA0 · October 2021 · Paper 2 · Q1 · 4 marks
4 marks
Examiner report
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. 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).
In an arithmetic series, the first term is 16 and the 21st term is 24.
Edexcel 9MA0 · October 2021 · Paper 2 · Q9 · 3 marks
3 marks
Examiner report
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 × ∑ 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.
Show that the infinite series from n = 2 to infinity of (3/4)^n cos(180n degrees) has sum 9/28.
Edexcel 9MA0 · October 2020 · Paper 1 · Q1 · 5 marks
5 marks
Examiner report
As hoped for, this proved to be a gentle and accessible start to the paper. The majority of prepared candidates scored all available marks in part (a). Errors, although rare, were mainly seen in the failure not to square the 8 of ( ) 2 8 . x Part (b) was found to be more demanding. Successful students substituted 1 32 x = into ( ) 1 2 1 8x + and noted that the result was 5 . 2 It then became a simple task of stating that you merely needed to substitute 1 32 x = into the expansion in (a) and multiply the result by 2. Concise explanations are still a weakness in the early years of this specification.
Consider the binomial expansion of (1+8x)^(1/2).
Edexcel 9MA0 · October 2020 · Paper 1 · Q5 · 6 marks
6 marks
Examiner report
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. 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.
A car has six forward gears. Its fastest speed is 28 km h^(-1) in first gear and 115 km h^(-1) in sixth gear.
Edexcel 9MA0 · October 2020 · Paper 1 · Q13 · 7 marks
7 marks
Examiner report
This question on sequences proved very discriminating, with weaker candidates scoring only the first mark. As with many questions on this paper however, there were some excellent and well written responses. In part (a) applying the iterative formula to find a2 was achieved by almost every candidate. A significant number then attempted to find a3 and also a4 in many cases but failed to simplify expressions for 2 4 2 k a = and 3 (2 2) 2 k k a k + = . Setting 4 1 a a = was often seen but fully correct proofs were rare. In part (b) only a very small minority of candidates gave a correct and sufficient reason why k cannot be equal to 1. Although a larger minority did state the sequence would be 2,2,2, … only a small number of these pointed out that this is not a periodic sequence with order 3. Part (c) could be done independently to (a) and (b), and many candidates took advantage of this. It was not uncommon to see the terms 2, 4, 1 − − in (c) where only the first mark in (a) had been scored. Methods of finding the sum to 80 terms generally revolved around the calculation ( ) ( ) ( ) 26 2 4 1) 2 4 × + −+ − + + − . Incorrect attempts involved uses of both arithmetic or geometric sum formula or ones involving an average of the terms.
A sequence a_1,a_2,a_3,... is defined by a_(n+1)=k(a_n+2)/a_n for positive integers n, where k is constant. The sequence has period of order 3 and a_1=2.
Edexcel 9MA0 · October 2020 · Paper 2 · Q4 · 3 marks
3 marks
Examiner report
This was generally well done, and most candidates scored full marks. A few candidates missed out the brackets around the 2x which usually led to a = 6. Several candidates made a mistake when finding the correct coefficient of x4 and scored no marks. This was either because the power of a was incorrect or the binomial coefficient was incorrect or occasionally completely missing. The most common error was having the wrong power of a, usually as a4 or sometimes a or a7 Several candidates used a7(1 + 2x/a)7 and errors with the power of a were also common with this method. Some candidates forgot to multiply by a7 and some forgot to divide the 2x by a.
In the binomial expansion of (a + 2x)^7, where a is a constant, the coefficient of x^4 is 15120.
Edexcel 9MA0 · October 2020 · Paper 2 · Q15 · 8 marks
8 marks
Examiner report
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. 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.
In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable. A geometric series has first term a and common ratio r, where a != 0 and r != 1. Its sum to n terms is S_n.
Edexcel 9MA0 · June 2019 · Paper 1 · Q4 · 6 marks
6 marks
Examiner report
(Mean Mark 3.3 out of 6) The vast majority of candidates did well in part (a), securing most or all of the marks available. They appreciated the need to rewrite the expression with a negative power and dealt confidently with the factorisation. The binomial expansion itself was also correctly executed and careful working ensured that there were relatively few slips in powers or signage. Common errors included bracketing errors seen with 2 4 x in the third term rather than 2 4 x failing to correctly combine the factored out 1 2 4 with the expansion 2 3 1 8 128 x x Part (b) caused more difficulty. In part (i), many candidates appeared to realise that the question was linked to the range of valid x values and the correct answer, 14 x , was selected most often. To gain this mark however, the correct answer and a valid reason needed to be given. Examples of valid reasons were ''the expansion is only valid for 4 x '' , ''the expansion is not valid for 14 x as 14 4 ''. In part (ii), only a small proportion of the candidates realised that the expansion is more accurate for values of x closest to 0. Many resorted to using the three values given and comparing the results to the exact value of 2 or else stating that the answer was 2 as it was the closest value to 2 .
Consider the binomial expansion of 1/sqrt(4-x). It may be used to approximate sqrt(2) by substituting x=-14, x=2 or x=-1/2 and then applying the corresponding scale factor.
Edexcel 9MA0 · June 2019 · Paper 1 · Q11 · 7 marks
7 marks
Examiner report
(Mean Mark 3.1 out of 7) 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. 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 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
A competitor runs a 20 km race. Each of the first 4 kilometres takes 6 minutes. From kilometre 5 onward, the model makes each kilometre take 5% longer than the previous kilometre.
Edexcel 9MA0 · June 2019 · Paper 2 · Q8 · 6 marks
6 marks
Examiner report
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 1 1 1 20 20 20 ... 2 2 2 and 5 5 5 3 4 5 log log log ... 2 3 4 gained a 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 solution and attempted to apply the formula . 1 a r Most used 1 2 r in this formula, although some used incorrect values such as 1 4 r or 1 . 2 r Disappointingly, a significant number of candidates used 4 r from the expression given in the question and applied it to the formula , 1 a r even though the formula only works when 1 1. r There were a number of correct strategies that candidates could use. The most successful strategy was to apply 4 1 1 20 , 2 2 a r to , 1 a r and many candidates who used this strategy achieved the correct answer of 2.5. Other complete strategies included applying 3 1 1 1 1 20 20 2 2 r r r r or 3 0 0 1 1 20 20 2 2 r r r r which led to (20 17.5) or (40 37.5) respectively. 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 to achieve 5 3 4 50 log ... . 2 3 49 Many candidates used cancelling to complete their proof correctly by writing 5 5 50 log log (25) 2. 2 The other method was to use the subtraction law of logarithms to give 48 5 5 1 (log ( 2) log ( 1)) n n n and list terms to give 5 5 5 5 5 5 (log 3 log 4 ...... log 50) (log 2 log 3 ...... log 49). Again many candidates used cancelling and completed their proof by writing 5 5 5 log (50) log (2) log (25) 2. A few candidates progressed by writing 5 5 log (3 4 ....... 50) log (2 3 ....... 49) and used factorials to give 5 5 5 5 50! 50! log log 49! log log (25) 2. 2 2(49!) Some candidates incorrectly assumed an arithmetic series and attempted to find its sum by using ( ) 2 n a l or [2 ( 1) ] 2 n a n d with 5 5 3 50 48, log , log 2 49 n a l and 5 5 4 3 log log . 3 2 d A few candidates incorrectly assumed a geometric series and 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.
This question concerns an infinite geometric sum and a telescoping logarithmic sum.
Edexcel 9MA0 · June 2018 · Paper 1 · Q11 · 10 marks
10 marks
Examiner report
This question on the binomial expansion was well attempted by many students, although showing sufficient detail to prove the given result in part (a) and providing a comprehensive explanation in part (b) proved difficult. In part (a), the majority of students started by attempting to expand 0.5 0.5 1 4 1 x x . As mentioned in previous years, it is good practice to give the unsimplified version (of each expansion) before attempting to find the simplified form. Those who did find the correct simplified forms of both expansions often lost the last two marks as many proceeded to just write down the given answer. This is a show that question and there was a requirement to show the six key terms that when collected together formed the given expression. Strangely there were a number of cases where the correct expansions were added, and in other cases divided, in an attempt to find the given answer. As with other ''explain'' questions on this paper, (b) was the worst attempted part of the question. Many merely pointed out that 1/2 was too big or could not be used to find 6, rather than focussing in on the range in values of x for which the expansion was valid. Part c was the most successful part of this question for many students with many gaining all three marks. Errors seen were mostly arithmetic, with the incorrect fraction 1183 968 commonly seen
Consider sqrt((1+4x)/(1-x)).
Edexcel 9MA0 · June 2018 · Paper 2 · Q4 · 7 marks
7 marks
Examiner report
This was an accessible question with part (ii) more successfully answered than part (i). In part (i), some students attempted to find 16 1 (3 5 2 ) r r r by applying the strategy 16 16 1 1 (3 16) 5 2 , r r r r while other students applied the strategy 16 16 1 1 (3 5 ) 2 . r r r r Most students applied the sum formula for an arithmetic series with 5, 5 a d to 16 1 (3 16) 5 r r to give 16 (3 16) (2(5) 15(5)) 2 while some applied 8, 5 a d to 16 1 (3 5 ) r r to give 16 (2(8) 15(5)). 2 A few students used 1 1 ( 1) 2 n r r n n to help them to evaluate 16 1 5 . r r Students were generally less successful with using the formula for the sum of a geometric series to find 16 1 2 . r r Common errors included assuming the first term, a, was 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 2 3 , u u correctly. Many deduced that they were working with an alternating sequence and found the correct exact value for 100 1 . r r u Some students attempted to round their answer, usually to 108.3. A few students attempted to find 100 1 r r u by using the formula for the sum of an arithmetic series or the formula for the sum of a geometric series.
A sequence u_1, u_2, u_3, ... is defined by u_(n+1) = 1/u_n, with u_1 = 2/3.
Edexcel 9MA0 · sample · Paper 2 · Q7 · 5 marks
5 marks
The binomial expansion of sqrt(4 - x), in ascending powers of x, begins 2 - x/4 + kx^2 + ... .
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · sample · Paper 2 · Q10 · 4 marks
4 marks
A convergent geometric series has common ratio r. Its sum to infinity is S_infinity and its sum to six terms is S_6. Given that S_infinity = (8/7)S_6, show that r = +/-1/sqrt(k), where k is an integer to be found.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 1 · Q2 · 8 marks
8 marks
Consider the binomial expansion of (4 + 5x)^(1/2).
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 1 · Q3 · 4 marks
4 marks
A sequence a_1, a_2, a_3, ... is defined by a_1 = 3 and a_(n+1) = (a_n - 3)/(a_n - 2), for positive integers n.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 1 · Q8 · 5 marks
5 marks
A farm harvested 2100 tonnes of wheat in 2017. The mass harvested in each subsequent year is expected to increase by 1.2% per year. Each year the first 2000 tonnes cost GBP 5.15 per tonne to harvest and any excess costs GBP 6.45 per tonne.
Pearson does not publish a direct link to this question paper.
Edexcel 9MA0 · specimen · Paper 2 · Q11 · 5 marks
5 marks
The second, third and fourth terms of an arithmetic sequence are 2k, 5k - 10 and 7k - 14 respectively, where k is a constant.
Pearson does not publish a direct link to this question paper.
Question extracts are shown for study alongside a link to the original Pearson paper. Reference-only rows carry series, paper, question number and marks.
Everything here stays free. If sequences and series is where the marks keep going, I teach it one to one from the mark schemes. Free intro call, then a free first lesson.