# IB Maths SL Sequence and Series Practice Problems Mr. W Name

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1 IB Maths SL Sequence and Series Practice Problems Mr. W Name Remember to show all necessary reasoning! Separate paper is probably best. 3b 3d is optional! 1. In an arithmetic sequence, u 1 = and u 3 = 8. Find d. Find u 0. Find S 0. () () (). In an arithmetic sequence u 1 = 7, u 0 = 64 and u n = Find the value of the common difference. Find the value of n. () (Total 5 marks) 3. Consider the arithmetic sequence 3, 9, 15,..., Write down the common difference. Find the number of terms in the sequence. Find the sum of the sequence. () 4. An arithmetic sequence, u 1, u, u 3,..., has d = 11 and u 7 = 63. Find u 1. () (i) Given that u n = 516, find the value of n. For this value of n, find S n. 5. The first three terms of an infinite geometric sequence are 3, 16 and 8. Write down the value of r. Find u 6. () Find the sum to infinity of this sequence. () (Total 5 marks) IB Questionbank Maths SL 1

2 6. The n th term of an arithmetic sequence is given by u n = 5 + n. Write down the common difference. (i) Given that the n th term of this sequence is 115, find the value of n. For this value of n, find the sum of the sequence. (5) 7. In an arithmetic series, the first term is 7 and the sum of the first 0 terms is 60. Find the common difference. Find the value of the 78 th term. () (Total 5 marks) In a geometric series, u 1 = and u4 = Find the value of r. Find the smallest value of n for which S n > 40. (Total 7 marks) 9. Expand 7 r= 4 r as the sum of four terms. (i) 30 Find the value of r= 4 r. Explain why =4 r r cannot be evaluated. (6) (Total 7 marks) 10. In an arithmetic sequence, S 40 = 1900 and u 40 = 106. Find the value of u 1 and of d. 11. Consider the arithmetic sequence, 5, 8, 11,... Find u 101. Find the value of n so that u n = 15. IB Questionbank Maths SL

3 1. Consider the infinite geometric sequence 3000, 1800, 1080, 648,. Find the common ratio. Find the 10 th term. Find the exact sum of the infinite sequence. () () () 13. Consider the infinite geometric sequence 3, 3(0.9), 3(0.9), 3(0.9) 3,. Write down the 10 th term of the sequence. Do not simplify your answer. Find the sum of the infinite sequence. (Total 5 marks) 14. In an arithmetic sequence u 1 = 37 and u 4 = 3. Find (i) the common difference; the first term. Find S 10. (Total 7 marks) 15. Let u n = 3 n. Write down the value of u 1, u, and u 3. 0 Find (3 n ). n= A theatre has 0 rows of seats. There are 15 seats in the first row, 17 seats in the second row, and each successive row of seats has two more seats in it than the previous row. Calculate the number of seats in the 0th row. Calculate the total number of seats. () 17. A sum of \$ 5000 is invested at a compound interest rate of 6.3 % per annum. Write down an expression for the value of the investment after n full years. What will be the value of the investment at the end of five years? IB Questionbank Maths SL 3

4 The value of the investment will exceed \$ after n full years. (i) Write down an inequality to represent this information. Calculate the minimum value of n. 18. Consider the infinite geometric sequence 5, 5, 1, 0.,. Find the common ratio. Find (i) the 10 th term; an expression for the n th term. Find the sum of the infinite sequence. 19. The first four terms of a sequence are 18, 54, 16, 486. Use all four terms to show that this is a geometric sequence. () (i) Find an expression for the n th term of this geometric sequence. If the n th term of the sequence is , find the value of n. 0. Write down the first three terms of the sequence u n = 3n, for n 1. Find (i) 0 n= n= 1 3n ; 3n. (5) 1. Consider the infinite geometric series For this series, find the common ratio, giving your answer as a fraction in its simplest form. Find the fifteenth term of this series. Find the exact value of the sum of the infinite series.. Consider the geometric sequence 3, 6, 1, 4,. (i) Write down the common ratio. Find the 15 th term. IB Questionbank Maths SL 4

5 Consider the sequence x 3, x +1, x + 8,. When x = 5, the sequence is geometric. (i) Write down the first three terms. Find the common ratio. () (d) Find the other value of x for which the sequence is geometric. For this value of x, find (i) the common ratio; the sum of the infinite sequence. (Total 1 marks) 3. Let S n be the sum of the first n terms of the arithmetic series Find (i) S 4 ; S 100. (PARTS B D ARE OPTIONAL CHALLENGES!!!!!) Let M = 1. (i) Find M. Show that M =. (5) It may now be assumed that M n = 1 n, for n 4. The sum Tn is defined by 1 (i) Write down M 4. T n = M 1 + M + M M n. Find T 4. (d) Using your results from part, find T 100. (Total 16 marks) IB Questionbank Maths SL 5

6 4. Clara organizes cans in triangular piles, where each row has one less can than the row below. For example, the pile of 15 cans shown has 5 cans in the bottom row and 4 cans in the row above it. A pile has 0 cans in the bottom row. Show that the pile contains 10 cans. There are 340 cans in a pile. How many cans are in the bottom row? (i) There are S cans and they are organized in a triangular pile with n cans in the bottom row. Show that n + n S = 0. Clara has 100 cans. Explain why she cannot organize them in a triangular pile. (6) (Total 14 marks) Worked Solutions 1. attempt to find d u3 u e.g. 1, 8 = + d d = 3 N correct substitution () e.g. u 0 = + (0 1)3, u 0 = u 0 = 59 N correct substitution () e.g. S 0 = 0 ( + 59), S0 = 0 ( ) S 0 = 610 N. evidence of choosing the formula for 0 th term e.g. u 0 = u d correct equation e.g = d, d = 19 d = 3 N 3 IB Questionbank Maths SL 6

7 correct substitution into formula for u n e.g = 7 + 3(n 1), 3709 = 3n + 4 n = 135 N1 [5] 3. common difference is 6 N1 evidence of appropriate approach e.g. u n = 1353 correct working e.g = 3 + (n 1)6, 6 n = 6 N evidence of correct substitution 6 ( ) 6 e.g. S 6 =, ( ) S 6 = (accept ) N1 4. evidence of equation for u 7 M1 e.g. 63 = u , u 7 = u 1 + (n 1) 11, 63 (11 6) u 1 = 3 N1 (i) correct equation e.g. 516 = 3 + (n 1) 11, 539 = (n 1) 11 n = 50 N1 correct substitution into sum formula 50( ) 50( ( 3) ) e.g. S 50 =, S 50 = S 50 = 135 (accept 1300) N r = = 3 N1 correct calculation or listing terms () e.g. 3,8, 3,... 4,, 1 u 6 = 1 N 3 evidence of correct substitution in S 3 3 e.g., S = 64 N1 [5] 6. d = N1 (i) 5 + n = 115 () n = 55 N IB Questionbank Maths SL 7

8 u 1 = 7 (may be seen in above) () correct substitution into formula for sum of arithmetic series e.g. S 55 = ( ), S 55 = ((7) + 54()), (5 + k) k = 1 S 55 = 3355 (accept 3360) N3 () 7. attempt to substitute into sum formula for AP (accept term formula) 0 0 e.g. S 0 = { ( 7) + 19d }, or ( 7 + u 0 ) setting up correct equation using sum formula 0 e.g. {( 7) + 19d} = 60 N correct substitution u 78 = () = 301 N [5] 8. evidence of substituting into formula for nth term of GP 1 3 e.g. u 4 = r setting up correct equation r 3 = 81 3 r = 3 N METHOD 1 setting up an inequality (accept an equation) 1 n 1 n (3 1) (1 3 ) e.g ; 81 n > > 40;3 > 6481 M1 evidence of solving e.g. graph, taking logs M1 n > () n = 8 N METHOD if n = 7, sum = ; if n = 8, sum = A n = 8 (is the smallest value) A N [7] 9. 7 r= 4 r 4 = (accept ) N1 (i) METHOD 1 recognizing a GP u 1 = 4, r =, n = 7 () correct substitution into formula for sum 4 7 ( 1) e.g. S 7 = 1 S 7 = N4 () IB Questionbank Maths SL 8

9 METHOD recognizing 30 = 30 3 r= 4 r= 1 r= 1 recognizing GP with u 1 =, r =, n = 30 correct substitution into formula for sum ( 30 1) S 30 = () 1 = r= 4 r = ( ) = N4 () valid reason (e.g. infinite GP, diverging series), and r 1 (accept r > 1) R1R1 N [7] 10. METHOD 1 substituting into formula for S 40 correct substitution 40( u 106) e.g = 1 + u 1 = 11 N substituting into formula for u 40 or S 40 correct substitution e.g. 106 = d, 1900 = 0( + 39d) d = 3 N METHOD substituting into formula for S 40 correct substitution e.g. 0(u d) = 1900 substituting into formula for u 40 correct substitution e.g. 106 = u d u 1 = 11, d = 3 NN 11. d = 3 () evidence of substitution into u n = a + (n 1) d e.g. u 101 = u 101 = 30 N3 correct approach e.g. 15 = + (n 1) 3 correct simplification () e.g. 150 = (n 1) 3, 50 = n 1, 15 = 1 + 3n n = 51 N 1. evidence of dividing two terms IB Questionbank Maths SL 9

10 e.g , r = 0.6 N evidence of substituting into the formula for the 10 th term e.g. u 10 = 3000( 0.6) 9 u 10 = 30. (accept the exact value ) N evidence of substituting into the formula for the infinite sum 3000 e. g. S = 1.6 S = 1875 N 13. u 10 = 3(0.9) 9 N1 recognizing r = 0.9 () correct substitution 3 e.g. S = S = 0.1 () S = 30 N3 14. (i) attempt to set up equations 37 = u 1 + 0d and 3 = u 1 + 3d 34 = 17d d = N [5] 3 = u 1 6 u 1 = 3 N1 u 10 = = 15 () 10 S 10 = (3 + ( 15)) M1 = 60 N [7] 15. u 1 = 1, u = 1, u 3 = 3 N3 Evidence of using appropriate formula M1 0 correct values S 0 = ( ) (= 10( 38)) S 0 = 360 N1 16. Recognizing an AP u 1 = 15 d = n = 0 () substituting into u 0 = 15 + (0 1) M1 = 53 (that is, 53 seats in the 0th row) N IB Questionbank Maths SL 10

11 0 0 Substituting into S 0 = ((15) + (0 1)) (or into ( )) M1 = 680 (that is, 680 seats in total) N (1.063) n N1 Value = \$ 5000(1.063) 5 (= \$ ) = \$ 6790 to 3 s.f. (accept \$ 6786, or \$ ) N1 (i) 5000(1.063) n > or (1.063) n > N1 Attempting to solve the inequality nlog(1.063) > log n > () 1 years N3 Note: Candidates are likely to use TABLE or LIST on a GDC to find n. A good way of communicating this is suggested below. Let y = x When x = 11, y = 1.958, when x = 1, y =.0816 x = 1 i.e. 1 years N3 () (0.) N1 5 (i) 9 1 u 10 = 5 5 = ,1.8 10, N n 1 1 u n = 5 N1 5 For attempting to use infinite sum formula for a GP = N 4 S = 31.5 ( = 31.3 to 3 s f ) 19. For taking three ratios of consecutive terms = = ( = 3) hence geometric AG N0 IB Questionbank Maths SL 11

12 (i) r = 3 () u n = 18 3 n 1 N For a valid attempt to solve 18 3 n 1 = eg trial and error, logs n = 11 N 0. 3, 6, 9 N1 (i) Evidence of using the sum of an AP M1 0 eg 3+ ( 0 1) 3 0 n = 1 3n = 630 N1 METHOD Correct calculation for 3n () 100 n = 1 eg ( ), Evidence of subtraction eg n = 1 3n = 1450 N METHOD Recognising that first term is 63, the number of terms is 80 ()() eg ( ), ( ) 100 n = 1 3n = 1450 N 1. For taking an appropriate ratio of consecutive terms r = 3 N For attempting to use the formula for the n th term of a GP u 15 = 1.39 N IB Questionbank Maths SL 1

13 For attempting to use infinite sum formula for a GP S = 115 N. (i) r = N1 u 15 = 3 ( ) 14 () = 4915 (accept 4900) N (i), 6, 18 N1 r = 3 N1 Setting up equation (or a sketch) M1 x + 1 x + 8 = x 3 x + 1 (or correct sketch with relevant information) x + x + 1 = x + x 4 () (d) (i) r = 1 x = 5 x = 5 or x = 5 x = 5 N Notes: If trial and error is used, work must be documented with several trials shown. Award full marks for a correct answer with this approach. If the work is not documented, award N for a correct answer. N1 For attempting to use infinite sum formula for a GP S = S = 16 N Note: Award M0A0 if candidates use a value of r where r > 1, or r < 1. [1] 3. (i) S 4 = 0 N1 u 1 =, d = () Attempting to use formula for S n M1 S 100 = N IB Questionbank Maths SL 13

14 (i) M 1 4 = A N For writing M 3 as M M or M M or M1 M 3 = A M 3 = 1 6 AG N0 (i) M = N1 T 4 1 = = 4 N3 (d) T 100 = = N3 [16] 4. Note: Throughout this question, the first and last terms are interchangeable. For recognizing the arithmetic sequence u 1 = 1, n = 0, u 0 = 0 (u 1 = 1, n = 0, d = 1) () Evidence of using sum of an AP M S S 0 = ( ) (or = ( )) S 0 = 10 AG N0 Let there be n cans in bottom row Evidence of using S n = n n n eg ( ) = 340, ( + ( n 1) ) = 340, ( n + ( n 1)( 1) ) = 340 n n + n 6480 = 0 n = 80 or n = 81 () n = 80 N IB Questionbank Maths SL 14

15 ( + n ) (i) Evidence of using S = 1 n S = n + n METHOD 1 n + n S = 0 AG N0 Substituting S = 100 ( eg n + n ) + n 400 = 0, 100 = 1 n EITHER n = 64.3, n = 65.3 Any valid reason which includes reference to integer being needed, R1 and pointing out that integer not possible here. R1 N1 eg n must be a (positive) integer, this equation does not have integer solutions. OR Discriminant = Valid reason which includes reference to integer being needed, R1 and pointing out that integer not possible here. R1 N1 eg this discriminant is not a perfect square, therefore no integer solution as needed. METHOD Trial and error S 64 = 080, S 65 = 145 Any valid reason which includes reference to integer being needed, R1 and pointing out that integer not possible here. R1 N1 [14] IB Questionbank Maths SL 15

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