Series Practice Problems 1. Find the sum of the arithmetic series Working:

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1 IB Math Standard Level Year Series Practice Series Practice Problems. Find the sum of the arithmetic series An arithmetic series has five terms. The first term is and the last term is 3. Find the sum of the series In an arithmetic sequence, the first term is 5 and the fourth term is 40. Find the second term Each day a runner trains for a 0 km race. On the first day she runs 000 m, and then increases the distance by 50 m on each subsequent day. (a) (b) On which day does she run a distance of 0 km in training? What is the total distance she will have run in training by the end of that day? Give your answer exactly. Answers: (a)... (b) The first three terms of an arithmetic sequence are 7, 9.5,. (a) What is the 4 st term of the sequence? (b) What is the sum of the first 0 terms of the sequence? Answers: (a)... (b)... Macintosh HD:Users:Shared:Dropbox:Desert:SL: - Algebra&Functions:Practice:SLA6SeriesPractice.docx on /30/5 at 9:33 AM Page of 4

2 IB Math Standard Level Year Series Practice 6. In an arithmetic sequence, the first term is, the fourth term is 6, and the n th term is 998. (a) Find the common difference d. (b) Find the value of n. Answers: (a)... (b)... (Total 6 marks) 7. Find the sum of the infinite geometric series $000 is invested at the beginning of each year for 0 years. The rate of interest is fixed at 7.5% per annum. Interest is compounded annually. Calculate, giving your answers to the nearest dollar (a) how much the first $000 is worth at the end of the ten years; (b) the total value of the investments at the end of the ten years. Answers: (a)... (b) Portable telephones are first sold in the country Cellmania in 990. During 990, the number of units sold is 60. In 99, the number of units sold is 40 and in 99, the number of units sold is 360. In 993 it was noticed that the annual sales formed a geometric sequence with first term 60, the nd and 3rd terms being 40 and 360 respectively. (a) What is the common ratio of this sequence? Assume that this trend in sales continues. (b) How many units will be sold during 00? (c) In what year does the number of units sold first exceed 5000? Between 990 and 99, the total number of units sold is 760. (d) What is the total number of units sold between 990 and 00? During this period, the total population of Cellmania remains approximately (e) Use this information to suggest a reason why the geometric growth in sales would not continue. () (3) (4) () () Macintosh HD:Users:Shared:Dropbox:Desert:SL: - Algebra&Functions:Practice:SLA6SeriesPractice.docx on /30/5 at 9:33 AM Page of 4

3 IB Math Standard Level Year Series Practice (Total marks) 0. The diagram shows a square ABCD of side 4 cm. The midpoints P, Q, R, S of the sides are joined to form a second square. (a) (i) Show that PQ = cm. (ii) Find the area of PQRS. (3) The midpoints W, X, Y, Z of the sides of PQRS are now joined to form a third square as shown. (b) (i) Write down the area of the third square, WXYZ. (ii) Show that the areas of ABCD, PQRS, and WXYZ form a geometric sequence. Find the common ratio of this sequence. The process of forming smaller and smaller squares (by joining the midpoints) is continued indefinitely. (c) (i) Find the area of the th square. (ii) Calculate the sum of the areas of all the squares.. Gwendolyn added the multiples of 3, from 3 to 3750 and found that = s. Calculate s. (3) (4) (Total 0 marks)... Macintosh HD:Users:Shared:Dropbox:Desert:SL: - Algebra&Functions:Practice:SLA6SeriesPractice.docx on /30/5 at 9:33 AM Page 3 of 4

4 IB Math Standard Level Year Series Practice (Total 6 marks). The diagrams below show the first four squares in a sequence of squares which are subdivided in half. The area of the shaded square A is. (a) (i) Find the area of square B and of square C. (ii) Show that the areas of squares A, B and C are in geometric progression. (iii) Write down the common ratio of the progression. (b) (i) Find the total area shaded in diagram. (ii) Find the total area shaded in the 8 th diagram of this sequence. Give your answer correct to six significant figures. (c) The dividing and shading process illustrated is continued indefinitely. Find the total area shaded. (5) (4) () (Total marks) Macintosh HD:Users:Shared:Dropbox:Desert:SL: - Algebra&Functions:Practice:SLA6SeriesPractice.docx on /30/5 at 9:33 AM Page 4 of 4

5 IB Math Standard Level Year Series Practice - MarkScheme Series Practice Problems - Markscheme (n )0 = 47 0(n ) = 400 n = 4 S 4 = ((7) + 40(0)) = 4(7 + 00) = 8897 OR S 4 = (7 + 47) = (434) = 8897 (C4). S 5 = { + 3} S 5 = 85 OR a =, a + 4d = 3 4d = 30 d = 7.5 S 5 = (4 + 4(7.5)) = (4 + 30) S 5 = 85 (C4) 3. a = 5 a + 3d = 40 (may be implied) d = T = 5 + = 6 or or 6.7 (3 sf) (C4) 4. (a) a = 000, a n = (n )50 = n = += She runs 0 km on the 37th day. (b) S 37 = 37 ( ) She has run a total of 03.5 km Macintosh HD:Users:Shared:Dropbox:Desert:SL: - Algebra&Functions:Practice:SLA6SeriesPractice.docx on /30/5 at 9:33 AM Page of 4

6 IB Math Standard Level Year Series Practice - MarkScheme 5. (a) u = 7, d =.5 u 4 = u + (n )d = 7 + (4 ).5 = 07 (C) (b) S 0 = n [u + (n )d] = 0 [(7) + (0 ).5] 0 64 = ( ) = 333 (C) 6. (a) u 4 = u l + 3d or 6 = +3d 7. S = ( ) 6 d = 3 = 6 (C3) (b) u n = u l + (n )6 or 998 = + (n l)6 n = = 00 (C3) 3 3 [6] = = 5 (C4) 8. (a) $ = $06 (nearest dollar) (C) (b) 000( ) ( ) ( ) =.075 = $508 (nearest dollar) (C3) 9. (a) r = = = 3 =.5 (b) 00 is the 3 th year. u 3 = 60(.5) 3 = 0759 (Accept 0760 or 0800.) 3 Macintosh HD:Users:Shared:Dropbox:Desert:SL: - Algebra&Functions:Practice:SLA6SeriesPractice.docx on /30/5 at 9:33 AM Page of 4

7 IB Math Standard Level Year Series Practice - MarkScheme (c) 5000 = 60(.5) n = (.5) n log = (n )log.5 n = = 8.49 n = th year 999 OR Using a gdc with u = 60, u k+ = u k, u 9 = 400, u 0 = 650 (M) 999 (G) 4 (d) S 3 = 60 = 6958 (Accept 6960 or 6000.) (e) Nearly everyone would have bought a portable telephone so there would be fewer people left wanting to buy one. (R) OR Sales would saturate. (R) 0. (a) (i) PQ = AP + AQ [] = + = 4 ( ) = cm (AG) (ii) Area of PQRS = ( ) ( ) = 8 cm 3 (b) (i) Side of third square = ( ) + ( ) = 4 = cm Area of third square = 4 cm (ii) st nd = 6 8 nd 3r nd = 8 4 Geometric progression, r = 8 6 = 4 8 = (c) (i) u = u r 0 = 6 0 = = ( = = 0.056, 3 sf) 64 Macintosh HD:Users:Shared:Dropbox:Desert:SL: - Algebra&Functions:Practice:SLA6SeriesPractice.docx on /30/5 at 9:33 AM Page 3 of 4

8 IB Math Standard Level Year Series Practice - MarkScheme (ii) S = u r = 6 = 3 4. Arithmetic sequence d = 3 (may be implied) n = 50 (A) S = ( ) or S = ( ) = (C6) [0] [6]. (a) (i) Area B = 6, area C = 64 (ii) 6 = = (Ratio is the same.) (R) 4 6 (iii) Common ratio = 4 (b) (i) Total area (S ) = = 5 6 = (= 0.35) (0.33, 3 sf) 5 (ii) Required area = S 8 = = (47...) = (6 sf) 4 Note: Accept result of adding together eight areas correctly. (c) Sum to infinity = 4 4 = 3 [] Macintosh HD:Users:Shared:Dropbox:Desert:SL: - Algebra&Functions:Practice:SLA6SeriesPractice.docx on /30/5 at 9:33 AM Page 4 of 4