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IEB practice · Paper I

IEB Sequences and series practice

13 newly generated questions across all 7 sequences and series topics the IEB examines in Paper I, 182 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

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Geometric sequences

7 questions · 92 marks

Paper I · Section A · 12 marks

QUESTION 1

Stage 1 of a pattern is a shaded equilateral triangle with sides of 20 cm. To make each next stage, every shaded triangle is divided into four equal triangles by joining the midpoints of its sides, and the middle one is removed. Stages 1, 2 and 3 are shown.

123
  1. (a)
    Show that the area of the triangle in stage 1 is 100√3 cm².
    (2)
  2. (b)
    How many shaded triangles are there in stage 7?
    (2)
  3. (c)
    Show that the areas removed in making stages 2, 3, 4, … form a geometric series, and calculate the total area removed in making stages 2 to 6, correct to two decimal places.
    (3)
  4. (d)
    If the pattern could go on for ever, what total area would be removed? What does this tell you about the shaded area?
    (2)
  5. (e)
    In which stage is the shaded area for the first time less than 30% of the area of stage 1?
    (3)

[12]

Show the answers

(a) Answer: 100√3 cm²

  1. area = ½ × 20 × 20 × sin 60°
  2. = ½ × 400 × √3/2 = 100√3 cm²

(b) Answer: 729

  1. stages 1, 2 and 3 have 1, 3 and 9: every shaded triangle leaves three
  2. stage 7: 3^6 = 729

(c) Answer: 132.10 cm²

  1. stage 2 removes ¼ of 100√3, which is 25√3 cm²
  2. each next stage removes 3 times as many triangles, each ¼ of the size: r = 3/4
  3. S_5 = 25√3(1 − 0.75^5)/(1 − 0.75) = 132.10 cm²

(d) Answer: 100√3 cm², all of stage 1: the shaded area tends to 0

  1. −1 < r = 0.75 < 1, so S∞ = 25√3/(1 − 0.75) = 100√3 cm²
  2. that is the whole area of stage 1, so the shaded area tends to 0

(e) Answer: stage 6

  1. 100√3(0.75)^(n − 1) < 0.3 × 100√3, so 0.75^(n − 1) < 0.3
  2. n − 1 > log 0.3/log 0.75 = 4.19 (log 0.75 < 0, so the inequality turns round)
  3. n − 1 = 5, so stage 6

Paper I · Section A · 12 marks

QUESTION 2

A square has sides of 16 cm. A second square is drawn by joining the midpoints of its sides, a third by joining the midpoints of the second's, and so on, as in the diagram.

  1. (a)
    Show that the side of the second square is 8√2 cm.
    (2)
  2. (b)
    Show that the areas of the squares form a geometric sequence, and write down its common ratio.
    (2)
  3. (c)
    Calculate the area of the sixth square.
    (2)
  4. (d)
    If the pattern carried on forever, what would the total area of all the squares be?
    (3)
  5. (e)
    Which square is the first to have an area of less than 1 cm^2?
    (3)

[12]

Show the answers

(a) Answer: 8√2 cm

  1. each side of the first square is cut into halves of 8 cm
  2. side^2 = 8^2 + 8^2 = 128, so the side is 8√2 cm

(b) Answer: r = 1/2

  1. 256; 128; 64; …
  2. each area is half the one before: r = 1/2

(c) Answer: 8 cm^2

  1. T_6 = 256(1/2)^5 = 8 cm^2

(d) Answer: 512 cm^2

  1. −1 < r < 1, so the series converges
  2. S = 256/(1 − 1/2) = 512 cm^2

(e) Answer: the tenth square

  1. 256(1/2)^(n − 1) < 1
  2. 2^(n − 1) > 256
  3. n − 1 > log 256/log 2 = 8.00
  4. n = 10

Paper I · Section A · 11 marks

QUESTION 3

The geometric sequence 25; 125; 625; … is given.

  1. (a)
    Write down the general term T_n of the geometric sequence.
    (1)
  2. (b)
    Show that log_5 T_1; log_5 T_2; log_5 T_3; … is an arithmetic sequence, and write down its first term and common difference.
    (3)
  3. (c)
    Calculate log_5 T_1 + log_5 T_2 + … + log_5 T_8.
    (3)
  4. (d)
    Hence write the product T_1 × T_2 × … × T_8 as a power of 5.
    (2)
  5. (e)
    Which term of the geometric sequence is equal to 5^12?
    (2)

[11]

Show the answers

(a) Answer: T_n = 25(5)^(n − 1)

  1. a = 25 and r = 125/25 = 5

(b) Answer: first term 2, common difference 1

  1. T_n = 5^2 × 5^(n − 1) = 5^(n + 1)
  2. log_5 T_n = n + 1: linear in n, so arithmetic
  3. log_5 T_(n+1) − log_5 T_n = 1 for every n

(c) Answer: 44

  1. an arithmetic series: a = 2, d = 1, n = 8
  2. S_8 = 8/2[2(2) + 7(1)] = 44

(d) Answer: 5^44

  1. log_5(T_1 × … × T_8) = log_5 T_1 + … + log_5 T_8 = 44
  2. T_1 × … × T_8 = 5^44

(e) Answer: T_11

  1. log_5 T_n = 12: n + 1 = 12
  2. n = 11

Paper I · Section A · 15 marks

QUESTION 4

A bamboo shoot breaks through the ground. It grows 4 cm on its first day, and on each day after that it grows 8% more than on the day before.

  1. (a)
    By how much does the shoot grow on day 8? Give your answer in cm, correct to two decimal places.
    (2)
  2. (b)
    How tall is the shoot at the end of the first 6 days? Give your answer to the nearest centimetre.
    (4)
  3. (c)
    On which day does the shoot first grow taller than 45 cm?
    (4)
  4. (d)
    3x − 33; 3x − 21 and 4x − 18 are the first three terms of a converging geometric sequence. Determine the value(s) of x.
    (5)

[15]

Show the answers

(a) Answer: 6.86 cm

  1. a geometric sequence with a = 4 and r = 1.08
  2. T_(8) = 4(1.08)^7 = 6.86

(b) Answer: 29 cm

  1. S_(6) = 4((1.08)^6 − 1)/(1.08 − 1)
  2. = 29 cm

(c) Answer: on day 9

  1. 4((1.08)^n − 1)/(1.08 − 1) ≥ 45
  2. (1.08)^n ≥ 1.9000
  3. n ≥ log 1.9000/log 1.08 = 8.34

(d) Answer: x = 3

  1. (3x − 21)^2 = (3x − 33)(4x − 18)
  2. x^2 − 20x + 51 = 0
  3. x = 3 or x = 17
  4. x = 17 gives r = 5/3: not between −1 and 1, so it does not converge
  5. x = 3 (r = 1/2)

Paper I · Section A · 13 marks

QUESTION 5

A patient takes one 80 mg tablet of a medicine at the same time every day. In the 24 hours between two tablets, the body removes 20% of the medicine that is in it.

  1. (a)
    Show that just after the third tablet there are 195.2 mg of the medicine in the body.
    (2)
  2. (b)
    Show that just after the nth tablet the amount of the medicine in the body is A_n = 400(1 − 0.8^n) mg.
    (3)
  3. (c)
    Calculate the amount of the medicine in the body just after tablet number 11, correct to two decimal places.
    (2)
  4. (d)
    If a tablet is taken every day without end, what amount does the medicine in the body approach just after a tablet? Explain why it never reaches this amount.
    (2)
  5. (e)
    The medicine works only once there are at least 155 mg of it in the body just after a tablet. After which tablet does this happen for the first time?
    (4)

[13]

Show the answers

(a) Answer: 195.2 mg

  1. just after tablet 1: 80 mg
  2. a day later 0.8 × 80 = 64 mg is left, so just after tablet 2: 64 + 80 = 144 mg
  3. just after tablet 3: 0.8 × 144 + 80 = 195.2 mg

(b) Answer: A_n = 400(1 − 0.8^n)

  1. A_n = 80 + 80(0.8) + 80(0.8)^2 + … + 80(0.8)^(n − 1)
  2. a geometric series with a = 80 and r = 0.8
  3. S_n = 80(1 − 0.8^n)/(1 − 0.8) = 400(1 − 0.8^n)

(c) Answer: 365.64 mg

  1. A_11 = 400(1 − 0.8^11)
  2. = 365.64 mg

(d) Answer: 400 mg

  1. −1 < r = 0.8 < 1, so the series converges: S∞ = 80/(1 − 0.8) = 400 mg
  2. 0.8^n > 0 for every n, so A_n = 400(1 − 0.8^n) is always less than 400 mg

(e) Answer: after tablet number 3

  1. 400(1 − 0.8^n) ≥ 155
  2. 0.8^n ≤ 0.6125
  3. n ≥ log 0.6125/log 0.8 = 2.20 (log 0.8 < 0, so the inequality turns round)

Paper I · Section A · 12 marks

QUESTION 6

A ball is dropped from a height of 10 m. After each bounce it rises to 1/2 of the height it fell from.

  1. (a)
    How high does the ball rise after the 4th bounce? Give your answer in metres, correct to two decimal places.
    (2)
  2. (b)
    Calculate the total vertical distance the ball has travelled when it hits the ground for the 7th time.
    (4)
  3. (c)
    If the ball bounces indefinitely, calculate the total vertical distance it travels.
    (3)
  4. (d)
    After which bounce does the ball first rise to a height of less than 25 cm?
    (3)

[12]

Show the answers

(a) Answer: 0.62 m

  1. a geometric sequence of heights with r = 1/2
  2. 10 × (1/2)^4 = 0.62 m

(b) Answer: 29.69 m

  1. down 10 m, then up and down again after each of 6 bounces
  2. 10 + 2 × 10(1/2)[1 − (1/2)^6]/(1 − 1/2)
  3. = 29.69 m

(c) Answer: 30.00 m

  1. −1 < r = 1/2 < 1, so the sum converges
  2. 10 + 2 × 10(1/2)/(1 − 1/2)
  3. = 30.00 m

(d) Answer: after bounce 6

  1. 10(1/2)^n < 0.25
  2. n > log(0.25/10)/log(1/2) = 5.32
  3. n = 6

Paper I · Section A · 17 marks

QUESTION 7
  1. (a)
    In a geometric sequence T_3 = 28 and T_6 = 224. Determine the first term and the common ratio.
    (4)
  2. (b)
    Calculate the sum of the first 9 terms.
    (2)
  3. (c)
    Determine the smallest value of n for which the sum of the first n terms is greater than 7 340 000.
    (5)
  4. (d)
    For which values of x will the series Σ_(n=1)^(∞) ((x + 1)/2)^n converge?
    (3)
  5. (e)
    Calculate the sum to infinity of the series when x = −2.
    (3)

[17]

Show the answers

(a) Answer: a = 7 and r = 2

  1. ar^2 = 28 and ar^5 = 224
  2. r^3 = 224/28 = 8
  3. r = 2
  4. a = 28/(2)^2 = 7

(b) Answer: S_(9) = 3577

  1. S_(9) = 7((2)^9 − 1)/(2 − 1)
  2. = 3577

(c) Answer: n = 20

  1. 7((2)^n − 1)/(2 − 1) > 7 340 000
  2. (2)^n > 1 048 572.4286
  3. n > log 1 048 572.4286/log 2
  4. n > 20.0000

(d) Answer: −3 < x < 1

  1. a geometric series with r = (x + 1)/2
  2. −1 < (x + 1)/2 < 1

(e) Answer: S_(∞) = −1/3

  1. r = (−2 + 1)/2 = −1/2
  2. a = T_1 = −1/2
  3. S_(∞) = a/(1 − r) = −1/3

Arithmetic sequences

3 questions · 47 marks

Paper I · Section A · 13 marks

QUESTION 8

The sum of the first n terms of a sequence is given by S_n = −2n^2 + 3n.

  1. (a)
    Determine the first three terms of the sequence.
    (3)
  2. (b)
    Show that the sequence is arithmetic, and determine T_n.
    (3)
  3. (c)
    Which term of the sequence is equal to −59?
    (2)
  4. (d)
    The sum of the first 6 terms of another arithmetic series is −72, and the sum of its first 14 terms is −280. Determine the first term and the common difference of this series.
    (5)

[13]

Show the answers

(a) Answer: 1; −3; −7

  1. S_1 = 1, S_2 = −2, S_3 = −9
  2. T_1 = S_1 = 1; T_2 = S_2 − S_1 = −3; T_3 = S_3 − S_2 = −7

(b) Answer: T_n = −4n + 5

  1. T_2 − T_1 = −4 and T_3 − T_2 = −4: a constant difference, so arithmetic
  2. T_n = 1 + (n − 1)(−4) = −4n + 5

(c) Answer: T_(16)

  1. −4n + 5 = −59
  2. n = 16

(d) Answer: a = −7 and d = −2

  1. S_6 = 6/2[2a + 5d] = −72, so 2a + 5d = −24
  2. S_14 = 14/2[2a + 13d] = −280, so 2a + 13d = −40
  3. subtracting: 8d = −16, so d = −2
  4. 2a − 10 = −24, so a = −7

Paper I · Section A · 17 marks

QUESTION 9
  1. (a)
    4x; 5x and 4x + 6 are the first three terms of an arithmetic sequence. Determine the value of x and the three terms.
    (3)
  2. (b)
    Which term of this sequence is equal to 78?
    (3)
  3. (c)
    How many terms of this sequence must be added to give a sum of 1116?
    (4)
  4. (d)
    A pattern of dots is built picture by picture. Pictures 1 to 4 have 10; 22; 40; 64 dots, and the numbers of dots form a quadratic sequence. How many dots are in picture 20?
    (4)
  5. (e)
    Which picture has 4450 dots?
    (3)

[17]

Show the answers

(a) Answer: x = 3; the terms are 12; 15; 18

  1. T_2 − T_1 = T_3 − T_2
  2. 2(5x) = (4x) + (4x + 6)
  3. 2x − 6 = 0
  4. x = 3

(b) Answer: T_(23)

  1. 12 + (n − 1)(3) = 78
  2. n − 1 = 22
  3. n = 23

(c) Answer: 24 terms

  1. n/2[2(12) + (n − 1)(3)] = 1116
  2. 3n^2 + 21n − 2232 = 0
  3. n = 24 (n is a natural number)

(d) Answer: 1264 dots

  1. second difference 6: a = 3
  2. T_n = 3n^2 + 3n + 4
  3. T_(20) = 1264

(e) Answer: picture 38

  1. 3n^2 + 3n + 4 = 4450
  2. 3n^2 + 3n − 4446 = 0
  3. n = 38 (n > 0)

Paper I · Section A · 17 marks

QUESTION 10

Consider the arithmetic series 30 + 28 + 26 + 24 + … − 100

  1. (a)
    Determine a formula for T_n, the general term of the series.
    (2)
  2. (b)
    Determine the number of terms in the series.
    (3)
  3. (c)
    Calculate the sum of the series.
    (3)
  4. (d)
    Write the series in sigma notation.
    (2)
  5. (e)
    The first five terms of a quadratic sequence are −10; −11; −8; −1; 10; … Determine T_n, the nth term.
    (4)
  6. (f)
    Determine whether 160 is a term of this quadratic sequence.
    (3)

[17]

Show the answers

(a) Answer: T_n = −2n + 32

  1. a = 30 and d = 28 − (30) = −2
  2. T_n = 30 + (n − 1)(−2) = −2n + 32

(b) Answer: 66 terms

  1. −2n + 32 = −100
  2. −2n = −132
  3. n = 66

(c) Answer: S_(66) = −2310

  1. S_n = n/2(a + l)
  2. S_(66) = 66/2(30 − 100)
  3. = −2310

(d) Answer: Σ_(n=1)^(66) (−2n + 32)

  1. the terms are T_n = −2n + 32, from n = 1 to n = 66

(e) Answer: T_n = 2n^2 − 7n − 5

  1. first differences: −1; 3; 7; 11
  2. second difference: 4, so 2a = 4 and a = 2
  3. 3a + b = −1, so b = −7
  4. a + b + c = −10, so c = −5

(f) Answer: Yes: it is T_(11)

  1. 2n^2 − 7n − 5 = 160
  2. 2n^2 − 7n − 165 = 0
  3. n = 11, a natural number

Sigma notation

1 questions · 14 marks

Paper I · Section A · 14 marks

QUESTION 11
  1. (a)
    Calculate Σ_(n=1)^(17) (5n + 4).
    (3)
  2. (b)
    Determine the value of p if Σ_(n=1)^(p) (5n + 5) = 675.
    (4)
  3. (c)
    Calculate Σ_(n=1)^(∞) (4(1/2)^(n − 1)).
    (3)
  4. (d)
    Determine the value of p if Σ_(n=1)^(p) (2(2)^(n − 1)) = 62.
    (4)

[14]

Show the answers

(a) Answer: 833

  1. an arithmetic series: a = 9, d = 5, n = 17
  2. S_(17) = 17/2[2(9) + (16)(5)] = 833

(b) Answer: p = 15

  1. a = 10, d = 5: p/2[2(10) + (p − 1)(5)] = 675
  2. 5p^2 + 15p − 1350 = 0
  3. p = 15 (p is a natural number)

(c) Answer: 8

  1. a geometric series with a = 4 and r = 1/2; −1 < r < 1
  2. S_(∞) = 4/(1 − (1/2)) = 8

(d) Answer: p = 5

  1. a geometric series with a = 2 and r = 2
  2. 2(2^p − 1)/(2 − 1) = 62
  3. 2^p = 32

Quadratic equations

2 questions · 29 marks

Paper I · Section A · 15 marks

QUESTION 12

An arts festival counts its visitors every day. On days 1, 2 and 3 it has 483, 522 and 555 visitors, and the numbers of visitors on days 1, 2, 3, … form a quadratic sequence.

  1. (a)
    How many visitors will the festival have on days 4 and 5?
    (2)
  2. (b)
    Determine an expression for T_n, the number of visitors on day n.
    (4)
  3. (c)
    On which day does the festival have the most visitors, and how many visitors does it have on that day?
    (3)
  4. (d)
    On which day does the festival first have fewer visitors than on day 1?
    (3)
  5. (e)
    Between which two consecutive days does the number of visitors fall by 81?
    (3)

[15]

Show the answers

(a) Answer: 582; 603

  1. first differences 39; 33; second difference −6
  2. 555 + (27) = 582; 582 + (21) = 603

(b) Answer: T_n = −3n^2 + 48n + 438

  1. 2a = −6, so a = −3
  2. 3a + b = 39, so b = 48
  3. a + b + c = 483, so c = 438

(c) Answer: day 8: 630 visitors

  1. the turning point of T is at n = −b/(2a) = −(48)/(2(−3)) = 8
  2. T_(8) = 630

(d) Answer: day 16

  1. −3n^2 + 48n + 438 < 483
  2. (n − 8)^2 > 49
  3. n − 8 > 7, so n > 15 (the sequence is symmetrical about day 8, so T_(15) = T_1)

(e) Answer: between day 21 and day 22

  1. the first differences form an arithmetic sequence: T_(k + 1) − T_k = −3(2k + 1) + 48
  2. −6k + 45 = −81
  3. k = 21: between day 21 and day 22

Paper I · Section A · 14 marks

QUESTION 13

The figures below are built from matchsticks: figure n is a rectangle of n rows of n + 1 small squares. Figures 1, 2 and 3 are shown.

123
  1. (a)
    How many matchsticks are needed for figure 5?
    (2)
  2. (b)
    Determine a formula for T_n, the number of matchsticks in figure n.
    (4)
  3. (c)
    Which figure needs exactly 647 matchsticks?
    (3)
  4. (d)
    Pieter has 2114 matchsticks. What is the largest figure Pieter can build, and how many matchsticks are left over?
    (3)
  5. (e)
    How many small squares are there in the figure that needs 199 matchsticks?
    (2)

[14]

Show the answers

(a) Answer: 71

  1. figures 1 to 4: 7; 17; 31; 49
  2. first differences 10; 14; 18, growing by 4
  3. 49 + 22 = 71

(b) Answer: T_n = 2n^2 + 4n + 1

  1. second difference 4: 2a = 4, so a = 2
  2. 3a + b = 10, so b = 4
  3. a + b + c = 7, so c = 1

(c) Answer: figure 17

  1. 2n^2 + 4n + 1 = 647
  2. 2n^2 + 4n − 646 = 0
  3. n = 17 (n > 0)

(d) Answer: figure 31, with 67 left over

  1. 2n^2 + 4n + 1 ≤ 2114
  2. n ≤ 31.52, so figure 31
  3. 2114 − T_31 = 2114 − 2047 = 67

(e) Answer: 90

  1. T_n = 199 gives n = 9
  2. figure 9 has 90 squares

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