Looking for a Tutor Near You?

Post Learning Requirement »
x
x

Direction

x

Ask a Question

x

Hire a Tutor

Note On Sequence And Series

Loading...

Published in: Mathematics
119 Views

Practice problems with Solutions

Mukul D / Dubai

6 years of teaching experience

Qualification: Bachelor's of Engineering

Teaches: Chemistry, Mathematics, SAT, ACT, Maths

Contact this Tutor
  1. Sequences 2008-2014 with MS 1.[4 marks] Find the value of k ifr=l 2a. [4 marks] The sum of the first 16 terms of an arithmetic sequence is 212 and the fifth term is 8. Find the first term and the common difference. 2b. [3 marks] Find the smallest value of n such that the sum of the first n terms is greater than 600. 3a. [2 marks] Each time a ball bounces, it reaches 95 % of the height reached on the previous bounce. Initially, it is dropped from a height of 4 metres. What height does the ball reach after its fourth bounce? 3b. [3 marks] How many times does the ball bounce before it no longer reaches a height of 1 metre? 3c. [3 marks] What is the total distance travelled by the ball? 4.[4 marks] Find the sum of all the multiples of 3 between 100 and 500. 5.[6 marks] A metal rod 1 metre long is cut into 10 pieces, the lengths of which form a geometric sequence. The length of the longest piece is 8 times the length of the shortest piece. Find, to the nearest millimetre, the length of the shortest piece. 6a. [3 marks] An arithmetic sequence has first term a and common difference d, d O . The3rd 4th and 7th terms of the arithmetic sequence are the first three terms of a geometric sequence. Show that 6b. [5 marks] Show that the 4th term of the geometric sequence is the 16th term of the arithmetic sequence. 7.[5 marks]
  2. In the arithmetic series with n term , it is given thatU4 = 7 and 22 Find the minimum value of n so that UI + 4- U34-... -Fun > 10 000 8a. [6 marks] (n+l)T The integralln is defined by In sin cldx, for n € N Show that O 8b. [4 marks] By letting Y = — MT , showthat n — 8c. [5 marks] I sin ald:r Hence determine the exact value of 0 9.[6 marks] The first terms of an arithmetic sequence are logsx log32X Find x if the sum of the first 20 terms of the sequence is equal to 100. 10.[6 marks] The mean of the first ten terms of an arithmetic sequence is 6. The mean of the first twenty terms of the arithmetic sequence is 16. Find the value of the 15th term of the sequence. 11.[7 marks] A geometric sequence has first term a, common ratio r and sum to infinity 76. A second geometric sequence has first term a, common ratio r3 and sum to infinity 36. Find r. 12a. [2 marks] The arithmetic sequence {un : n Z } has first term ul {v : n e Z + } has first term VI The geometric sequence n Find an expression for — Vn in terms of n. 12b. [3 marks] Determine the set of values of n for which Un > tin. and common difference d = 1.5 3 and common ratio r = 1.2. 12c. [1 mark] Determine the greatest value of un 13a. [4 marks] — Vn. Give your answer correct to four significant figures. (i) Express the sum of the first n positive odd integers using sigma notation. 2
  3. 2 (ii) Show that the sum stated above is n (iii) Deduce the value of the difference between the sum of the first 47 positive odd integers and the sum of the first 14 positive odd integers. 13b. [7 marks] A number of distinct points are marked on the circumference of a circle, forming a polygon. Diagonals are drawn by joining all pairs of non-adjacent points. (i) Show on a diagram all diagonals if there are 5 points. if there are n points, where n > 2. (ii) Show that the number of diagonals is 2 (iii) Given that there are more than one million diagonals, determine the least number of points for which this is possible. 13c. [8 marks] The random variable X P) has mean 4 and variance 3. (i) Determine n and p. (ii) Find the probability that in a single experiment the outcome is 1 or 3. 14a. [4 marks] Find the set of values of x for which the series 14b. [2 marks] Hence find the sum in terms of x. 15a. [9 marks] has a finite sum. In an arithmetic sequence the first term is 8 and the common difference is 7. If the sum of the first 2n terms is equal to the sum of the next n terms, find n. 15b. [7 marks] Ifal' (12' a.3' • • • are terms of a geometric sequence with common ratior I, show that (al — a2)2 + (a2 — (a3 — 04) 2 (an — an+1)2 — 16a. [1 mark] Show that 16b. [2 marks] Consider the geometric series Write down the common ratio, z, of the series, and show that 3. 3
  4. 16c. [2 marks] Find an expression for the sum to infinity of this series. 16d. [8 marks] 9 sin 0 sin 9+ {sin20+ {sin30+ Hence, show that — 10—6cos9. 17a. [2 marks] 81 A geometric sequence ul , U2 , u.3 , has UI = 27 and a sum to infinity of T. Find the common ratio of the geometric sequence. 17b. [5 marks] An arithmetic sequence VI , V2 , V3 , is such thatV2 = t-12 and V4 = Evn>0 Find the greatest value of N such that 18.[17 marks] A geometric sequence {un}, with complex terms, is defined by un+l = (l i)un and ul 3. (a) Find the fourth term of the sequence, giving your answer in the form + Y e R (b) Find the sum of the first 20 terms of {un}, giving your answer in the form a X (l + 2m) where a e C and m e Z are to be determined. A second sequence {vn} is defined byVn = UnUn+k, k N (c) (i) Show that {vn} is a geometric sequence. (ii) State the first term. (iii) Show that the common ratio is independent of k. A third sequence wn} is defined byWn = lun — tin +11. (d) (i) Show that {wn} is a geometric sequence. (ii) State the geometrical significance of this result with reference to points on the complex plane. 19.[7 marks] The first three terms of a geometric sequence are Sin X, sin and 4 sinæcos C, 2 (a) Find the common ratio r. (b) Find the set of values of x for which the geometric series 4- Sin2æ 4- 4Sinæcos a: + . converges. 4
  5. Consider C = arccos (c) Show that the sum to infinity of this series is 2 . 20.[6 marks] (a) (i) Find the sum of all integers, between 10 and 200, which are divisible by 7. (ii) Express the above sum using sigma notation. An arithmetic sequence has first term 1000 and common difference of —6 . The sum of the first n terms of this sequence is negative. (b) Find the least value of n. 21.[7 marks] The sum of the first two terms of a geometric series is 10 and the sum of the first four terms is 30. (a) Show that the common ratio r satisfiesr2 2. (b) Given r — (i) find the first term; (ii) find the sum of the first ten terms. 22.[7 marks] The fourth term in an arithmetic sequence is 34 and the tenth term is 76. (a) Find the first term and the common difference. (b) The sum of the first n terms exceeds 5000. Find the least possible value of n. 5
  6. Sequences 2008-2014 MS 1.[4 marks] Markscheme (Al) (Al) -j Ml [4 marks] Examiners report The question was well done generally. Those that did make mistakes on the question usually had the first term wrong, but did understand to use the formula for an infinite geometric series. 2a. [4 marks] Markscheme 212— pa + 15d) 16a+ 120d) Al nth term is a -F (n — l)d 8 Al solving simultaneously: (Ml) 1.5, [4 marks] Examiners report This proved to be a good start to the paper for most candidates. The vast majority made a meaningful attempt at this question with many gaining the correct answers. Candidates who lost marks usually did so because of mistakes in the working. In part (b) the most efficient way of gaining the answer was to use the calculator once the initial inequality was set up. A small number of candidates spent valuable time unnecessarily manipulating the algebra before moving to the calculator. 2b. [3 marks] Markscheme [4+ 1.5(n — 1)] > 600 (Ml) 3n2 +5n - 2400 > O (Al) 6
  7. • n > 27.4..., (n < —29.1...) Note: Do not penalize improper use of inequalities. n 28 [3 marks] Examiners report This proved to be a good start to the paper for most candidates. The vast majority made a meaningful attempt at this question with many gaining the correct answers. Candidates who lost marks usually did so because of mistakes in the working. In part (b) the most efficient way of gaining the answer was to use the calculator once the initial inequality was set up. A small number of candidates spent valuable time unnecessarily manipulating the algebra before moving to the calculator. 3a. [2 marks] Markscheme height 4 X 0.954 (Al) = 3.26 (metres) Al [2 marks] Examiners report The majority of candidates were able to start this question and gain some marks, but only better candidates gained full marks. In part (a) the common error was to assume the wrong number of bounces and in part (b) many candidates lost marks due to rounding the inequality in the wrong direction. Part (c) was found difficult with only a limited number recognising the need for the sum to infinity of a geometric sequence and many of those did not recognise how to link the sum to infinity to the total distance travelled. 3b. [3 marks] Markscheme 4 X 0.95n < 1 (Ml) 0.95n < 0.25 In O.25 In 0.95 (Al) > 27.0 Note: Do not penalize improper use of inequalities. n 28 Al Note: If candidates have used n — 1 rather than n throughout penalise in part (a) and treat as follow through in parts (b) and (c). [3 marks] 7
  8. Examiners report The majority of candidates were able to start this question and gain some marks, but only better candidates gained full marks. In part (a) the common error was to assume the wrong number of bounces and in part (b) many candidates lost marks due to rounding the inequality in the wrong direction. Part (c) was found difficult with only a limited number recognising the need for the sum to infinity of a geometric sequence and many of those did not recognise how to link the sum to infinity to the total distance travelled. 3c. [3 marks] Markscheme METHOD 1 recognition of geometric series with sum to infinity, first term of4 X 0.95 and common ratio 0.95 Ml recognition of the need to double this series and to add 4 MI total distance travelled is 2 + 4 156 (metres) Al [3 marks] Note: If candidates have used n — 1 rather than n throughout penalise in part (a) and treat as follow through in parts (b) and (c). METHOD 2 recognition of a geometric series with sum to infinity, first term of 4 and common ratio 0.95 MI recognition of the need to double this series and to subtract 4 MI . 2 ( 4 ) —4 = 156 (metres) Al total distance travelled IS 1-0.95 [3 marks] Examiners report The majority of candidates were able to start this question and gain some marks, but only better candidates gained full marks. In part (a) the common error was to assume the wrong number of bounces and in part (b) many candidates lost marks due to rounding the inequality in the wrong direction. Part (c) was found difficult with only a limited number recognising the need for the sum to infinity of a geometric sequence and many of those did not recognise how to link the sum to infinity to the total distance travelled. 4.[4 marks] Markscheme METHOD 1 102 + 105 + + 498 (Ml) so number of terms = 133 (Al) 8
  9. EITHER = (2 x 102+ 132 x 3) (Ml) - 39900 Al OR 133 (102 + 498) x T (Ml) - 39900 Al OR 166 3n - 39900 Al METHOD 2 500 +3 = 166.666... and 100 3 = 33.333... 166 33 102 + 105 +498 — 3n 166 E 3n = 41583 (Al) 33 E 1683 (Al) the sum is 39900 Al [4 marks] Examiners report Most candidates got full marks in this question. Some mistakes were detected when trying to find the number ofterms of the arithmetic sequence, namely the use of the incorrect value n = 132 ; a few interpreted the question as the sum ofmultiples between the 100th and 500th terms. Occasional application of geometric series was attempted. 5.[6 marks] Markscheme the pieces have lengths a, ar, . 80 = arg (or 8 = TO) Al r — 1.259922... Al , arg (Ml) 9
  10. a (ora = 1000) Ml r 1 —00286 (Al) a = 29 mm (accept 0.029 m or any correct answer regardless the units) Al [6 marks] Examiners report This question was generally well done by most candidates. Some candidates recurred to a diagram to comprehend the nature of the problem but a few thought it was an arithmetic sequence. A surprising number of candidates missed earning the final Al mark because they did not read the question instructions fully and missed the accuracy instruction to give the answer correct to the nearest mm. 6a. [3 marks] Markscheme let the first three terms of the geometric sequence be given by ul , ul r , ul r2 . ul = a +2d tllr = a +3dand = a + 6d (Ml) a+6d a+3d a+3d a+2d Al a2 + 8ad + 12d2 2a +3d=O [3 marks] Examiners report This question was done well by many students. Those who did not do it well often became involved in convoluted algebraic processes that complicated matters significantly. There were a number of different approaches taken which were valid. 6b. [5 marks] Markscheme d = (ult.2 = ul = 2 Ml r=3A1 ulP3 _ geometric 4 term 27d arithmetic 16th term Ml 10
  11. 27d T Al Note: Accept alternative methods. [3 marks] Examiners report This question was done well by many students. Those who did not do it well often became involved in convoluted algebraic processes that complicated matters significantly. There were a number of different approaches taken which were valid. 7.[5 marks] Markscheme = ul +3d= +8d = 22 Note: 5d = 15 gains both above marks - (-4 + (n - 1)3) > 10000 Ml n 83M [5 marks] Examiners report This question was well answered by most candidates. A few did not realise that the answer had to be an integer. 8a. [6 marks] Markscheme 10 = for e— sin xd:r Ml 10 = e-tlsinxldc Note: Award MI for Attempt at integration by parts, even if inappropriate modulus signs are present. MI [ -x -for e-c cos :rda — —[e—x sinc]o — cosxdæ Al = —[e-x cosæ]or — [e—t sinc]å — e-t sin Sina + cosæ]or — e-c sin Cda Al = —[e-r cosa]å — [e —x sinc]å — 10 or —[e-x sinc + e-z cosæ]o Note: Do not penalise absence of limits at this stage 11 — 10 Ml
  12. 10 — (1 + AG Note: If modulus signs are used around cos x , award no accuracy marks but do not penalise modulus signs around sin x . [6 marks] Examiners report Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in O which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen. 8b. [4 marks] Markscheme (n+l)r e — x sin Attempt to use the substitution Y = — MI (putting Y = — dc and (n + 1)7T] [0, In = f r sin(y + nn)ldy Al sin ydYA1 e-nTIo AG [4 marks] Examiners report Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in O which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen. 8c. [5 marks] Markscheme (Al) 12
  13. the term is an infinite geometric series with common ratio e —r (Ml) therefore 10 sincldx — — I-e-t (Al) 2(er-1) Al [5 marks] Examiners report Part (a) is essentially core work requiring repeated integration by parts and many candidates realised that. However, some candidates left the modulus signs in O which invalidated their work. In parts (b) and (c) it was clear that very few candidates had a complete understanding of the significance of the modulus sign and what conditions were necessary for it to be dropped. Overall, attempts at (b) and (c) were disappointing with few correct solutions seen. 9.[6 marks] Markscheme METHOD 1 logs:r — log2X (MI) log28 (Ml) Note: Award this MI for a correct change of base anywhere in the question. — 24 — 16M 20 2 x —L +19 x log2X — log2Z 400 100 = log.2X 4 METHOD 2 logar 20th term = 20 100 = log2 20 100 = log. Al log2 239 log-2 Ml Ml Ml (Al) 13
  14. 400 100 = log.2a: = 4 = 2 METHOD 3 Note: Award this MI for a correct change of base anywhere in the question. — 24 = 16M log2X log2 logsx log32 log,s log,32 log2X log2X = 16M 10428 x log,128 log2 z Note: Award this MI for a correct change of base anywhere in the question. = (1+3+54...) Al — (2 + 38)) log.2Z 2 100 — — log2C log2a; = 4 [6 marks] Examiners report There were plenty of good answers to this question. Those who realised they needed to make each log have the same base (and a great variety of bases were chosen) managed the question successfully. 10.[6 marks] Markscheme METHOD 1 5(2a + 9d) = 60 (or 2a + 9d = 12) MIAI 10(2a + 19d) 320 (or + 19d 32) Al solve simultaneously to obtain MI a = -3, d=2A1 the 15th term is -3+ 14 X 2 25 Al Note: FT the final Al on the values found in the penultimate line. METHOD 2 with an AP the mean of an even number of consecutive terms equals the mean of the middle terms (Ml) 14
  15. = 16 (or + all = 32) Al = 6 (ora5 -F (16 = 12) Al (110 — a5 + all — (16 = 20M1 5d+5d = 20 — —3 (or 15) Al d 2 and a — 5 or al() —3+ 14 x 2 —25 (or 5+ 10 x 2 = 25 or 15+5 x 2 — 25) Al the 15th term is Note: FTthe final Al on the values found in the penultimate line. [6 marks] Examiners report Many candidates had difficulties with this question with the given information often translated into incorrect equations. 11.[7 marks] Markscheme -c — 76 for the first series I—r Al a —36 for the second series Al 76(1 -r) 36 attempt to eliminate a e.g. I—ra Ml simplify and obtain 9r2 + 9r — 10 O (MI)AI Note: Only award the Ml if a quadratic is seen. _ 12 30 — and — (Al) obtain — = 0.666..•) Al = 18 \ — Note: Award AO if the extra value of r is given in the final answer. Total [7 marks] Examiners report Almost all candidates obtained the cubic equation satisfied by the common ratio of the first sequence, but few were able to find its roots. One of the roots was r = 1. 12a. [2 marks] Markscheme 15
  16. - 1) x 1.5 -3 x 1.2n-1 (— — 1.5n+O.1 -3 x 1.2n 1 un — vn = 1.6 + (n [2 marks] Examiners report ) AIAI In part (a), most candidates were able to express Un and Vn correctly and hence obtain a correct expression for — Vn. Some candidates made careless algebraic errors when unnecessarily simplifying Un while other candidates incorrectly stated as (12) n 12b. [3 marks] Markscheme attempting to solve Un > Vn numerically or graphically. (MI) n=2.621 . . . , 9.695...(A1) s03< n < 9 Al [3 marks] Examiners report In parts (b) and (c), most candidates treated n as a continuous variable rather than as a discrete variable. Candidates should be aware that a GDC's table feature can be extremely useful when attempting such question types. 12c. [1 mark] Markscheme The greatest value of un Note: Do not accept 1.64. [1 mark] Examiners report — vn is 1.642. Al In parts (b) and (c), most candidates treated n as a continuous variable rather than as a discrete variable. Candidates should be aware that a GDC's table feature can be extremely useful when attempting such question types. In part (c), a number of candidates attempted to find the maximum value of n rather than attempting to find the maximum value of un — Vn. 13a. [4 marks] Markscheme E (2k - 1) (or equivalent) Al (2n — 1) Note: Award AO for or equivalent. 16
  17. (ii) EITHER n(n+l) OR OR THEN MIAI — 1)2) (using Sn — — 1) (using Sn — — (2u1 + (n — l)d)) MIAI — (UI -Fun)) MIAI (iii) 47 [4 marks] 142 — 2013M Examiners report In part (a) (i), a large number of candidates were unable to correctly use sigma notation to express the sum of the first n positive odd integers. Common errors included summing 2n — I from 1 to n and specifying sums with incorrect limits. Parts (a) (ii) and (iii) were generally well done. 13b. [7 marks] Markscheme (i) EITHER a pentagon and five diagonals Al OR five diagonals (circle optional) Al (ii) Each point joins to n - 3 other points. Al a correct argument forn n — 3) RI n(n—3) a correct argument for 2 RI 3) > 1000000 (iii) attempting to solve 2 for n. (Ml) n > 1415.7 (Al) n 1416M [7 marks] Examiners report 17
  18. Parts (b) (i) and (iii) were generally well done. In part (b) (iii), many candidates unnecessarily simplified their quadratic when direct GDC use could have been employed. A few candidates gave n > 1416 as their final answer. While some candidates displayed sound reasoning in part (b) (ii), many candidates unfortunately adopted a 'proof by example' approach. 13c. [8 marks] Markscheme (i) np = 4 and npq = 3 (Al) attempting to solve for n and p (MI) n 16 andP— Al (ii) X B(16, 0.25) (Al) 16 1 16 3 (Al) - 0.261 Al [8 marks] Examiners report Part (c) was generally well done. In part (c) (ii), some candidates multiplied the two probabilities rather than adding the two probabilities. 14a. [4 marks] Markscheme for the series to have a finite sum, RI (sketch from gdc or algebraic method) MI exists when — 3 AIM Note: Award Al for bounds and Al for strict inequalities. [4 marks] Examiners report A large number of candidates omitted the absolute value sign in the inequality in (a), or the use of the correct double inequality. Among candidates who had the correct statement, those who used their GDC were the most successful. The algebraic solution of the inequality was difficult for some 18
  19. candidates. In (b), quite a number of candidates found the sum of the first n terms of the geometric series, rather than the infinite sum of the series. 14b. [2 marks] Markscheme MIAI [2 marks] Examiners report A large number of candidates omitted the absolute value sign in the inequality in (a), or the use of the correct double inequality. Among candidates who had the correct statement, those who used their GDC were the most successful. The algebraic solution of the inequality was difficult for some candidates. In (b), quite a number of candidates found the sum of the first n terms of the geometric series, rather than the infinite sum of the series. 15a. [9 marks] Markscheme — (2(8) -F (2n = n (16 -F Al — (16 + — 4 Al S2n solve 2n (16 (4 2 (16 = S3n = S3n Ml (16 + Al (16 + (gdc or algebraic solution) (Ml) n = 63 A2 [9 marks] Examiners report Many candidates were able to solve (a) successfully. A few candidates failed to understand the relationship between 2n and S3n, and hence did not obtain the correct equation. (b) was answered poorly by a large number of candidates. There was significant difficulty in forming correct general statements, and a general lack of rigor in providing justification. 15b. [7 marks] 19
  20. Markscheme (al — 02) 2 — a3)2 + — = (al — al + (al r — alr2)2 + (arr2 — airs) -F [al (1 — + [a IT.(I _ + [al — 2 Note: This Al is for the expression for the last term. • MIAI . + [alt.n-i(l (Al) = af(l — -F afr2(I — + — + + afr2n-2(1 — Al [7 marks] Examiners report Many candidates were able to solve (a) successfully. A few candidates failed to understand the relationship between 2n and S3n, and hence did not obtain the correct equation. (b) was answered poorly by a large number of candidates. There was significant difficulty in forming correct general statements, and a general lack of rigor in providing justification. 16a. [1 mark] Markscheme lei91 IcosO+isinOl) = [1 mark] Examiners report cos29 sm 9 MIAG Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates. 16b. [2 marks] Markscheme Z L ei0 3 Al Izl låeigl_L 3 AIAG [2 marks] 20
  21. Examiners report Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates. 16c. [2 marks] Markscheme 1-9" (MI)AI [2 marks] Examiners report Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates. 16d. [8 marks] Markscheme EITHER — Al 1 — \ cos sin — cos 9— L i sin 9) (I—L c:os 9) MIAI — cos 9+\isin 9 — cos O) Al — cos sin 9 1—; 9+ Al OR I ei9) M IAI — Al 21
  22. I— k (cos sin O) Al THEN taking imaginary parts on both sides sin O + sin 20 -IL ... = sin O 10 2 — cos e sine + sin 20 + . [8 marks] Examiners report — sin — cos 9 MIAIAI 9 sin 9 — 10—6cos9 AG Parts (b) and (c) were answered fairly well by quite a few candidates. In (a) many candidates failed to write the formula for the modulus of a complex number. (c) proved inaccessible for a large number of candidates. The algebraic manipulation required and the recognition of the imaginary and real parts in order to arrive at the necessary relationship were challenging for many candidates. 17a. [2 marks] Markscheme 81 _ 27 = 27T [2 marks] — I-r Ml — i Al Examiners report Part (a) was well done by most candidates. However (b) caused difficulty to most candidates. Although a number of different approaches were seen, just a small number of candidates obtained full marks for this question. 17b. [5 marks] Markscheme 9v4 — 12d - — 13 (Al) 8 d — —4 (Al)V1 (2 x 13 -4(N-1)) > o (accept equality) Ml (30 - 4N) > O N (15 — 2N) > O N < 7.5 (Ml) Note: 13 9+5 I or equivalent receives full marks. 22
  23. [5 marks] Examiners report Part (a) was well done by most candidates. However (b) caused difficulty to most candidates. Although a number of different approaches were seen, just a small number of candidates obtained full marks for this question. 18.[17 marks] Markscheme = _6 + 6iA1 [3 marks] ((1+i)20-1 (Ml) 10-1) (Ml) Note: Only one of the two MIS can be implied. Other algebraic methods may be seen. (Al) = 3i (210 + 1) Al [4 marks] (c) (i) METHOD 1 = (3(1 + i)n-l) (3(1 + i) n—l+k + + i)2n-2 Al = +02) n 1 + 1) this is the general term of a geometrical sequence RIAG Notes: Do not accept the statement that the product of terms in a geometric sequence is also geometric unless justified further. If the final expression forth is + + i)2n award MIAIRO. METHOD 2 Vn+l 23
  24. this is a constant, hence sequence is geometric RIAG Note: Do not allow methods that do not consider the general term. (ii) 9(1 + Al (iii) common ratio is (l + ( = 2i) (which is independent of k) Al [5 marks] (d) (i) METHOD 1 wn 13(1 + t) • '1-1 -3(1 +i)nlM1 = 311 -kiln-I Il — (I + i)lM1 311 Al 366) this is the general term for a geometric sequence RIAG METHOD 2 ton = — (1 + MI = lunl I—il = lunlA1 — 13(1 + 1 this is the general term for a geometric sequence RIAG Note: Do not allow methods that do not consider the general term. (ii) distance between successive points representing Un in the complex plane forms a geometric sequence RI Note: Various possibilities but must mention distance between successive points. [5 marks] Total [17 marks} Examiners report 24
  25. 19.[7 marks] Markscheme (a) sinx, sin2æ and 4sin ccos2x 2 sin a: cos a: 2 cost Al sin 2a: Note: Accept sin x . [1 mark] (b) EITHER Ir < 1 * 12cosxl < IMI OR —1 < r < 1-9 THEN 0 < cost < [3 marks] —1 < 2COSX < IMI for AMI 1 —2 cos arccos AIAI Note: Award Al for correct numerator and Al for correct denominator. [3 marks] Total [7 marks] Examiners report [N/A] 20.[6 marks] Markscheme 25
  26. (a) (i) n 27 (Al) METHOD 1 14 196 S27 x 27 (Ml) 2835M METHOD 2 sn = (2 x 14 +26 x 7) (Ml) = 2835M METHOD 3 27 S27 7 + 7n (Ml) = 2835M 27 or equivalent Al 28 Note: Accept [4 marks] n (2000 — — 1)) < 0 (Ml) n > 334.333 n 335 Al Note: Accept working with equalities. [2 marks] Total [6 marks] Examiners report [N/A] 21.[7 marks] Markscheme (a) METHOD 1 a + ar = 10 Al = 30 Al 26
  27. Examiners report This question was invariably answered very well. Candidates showed some skill in algebraic manipulation to derive the given answer in part a). Poor attempts at part b) were a rarity, though the final mark was sometimes lost after a correctly substituted equation was seen but with little follow-up work. 22.[7 marks] Markscheme (a) METHOD 1 a + ar = 10 ar2 + arg = IOr2 orar2 + arg 10 + 10r2 = 30 orr2 (a + CT) = 20 Al METHOD 2 = 10 d- 7M = 20 Ml and I—r Ml leading to either I -f- 7.2 [4 marks] (b) (i) a + — 10 10 = 30 MIAI = 3 (or r4 — 3r2 + 2 — O) Al — = 10 (v'ä — 1) Al 10 SIO = 1+0 (ii) 310M [3 marks] Total [7 marks] 0-1 10 x 31) Ml 34 76 a = 13M METHOD 2 76 = 34 + 6d (Ml) = a + 9d (Ml) 27
  28. d = 7M 34 —a + 21 a = 13M [3 marks] (26 + 7(n — 1)) > 5000 n > 36.463... Note: Award MIAIAI for using either an equation, a graphical approach or a numerical approach. n 37M [4 marks] Total [7 marks] Examiners report Both parts were very well done. In part (a), a few candidates made a careless algebraic error when attempting to find the value of a or d. In part (b), a few candidates attem ted to find the value of n for which Un > 5000 . Some candidates S — [ul + (n — l)d] used the incorrect formula n . A number of candidates unnecessarily attempted to simplify SR. Most successful candidates in part (b) adopted a graphical approach and communicated their solution effectively. A few candidates did not state their value of n as an integer. 28