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IEB practice · Paper I

IEB Sequences and series practice

4 newly generated questions across all 7 sequences and series topics the IEB examines in Paper I, 64 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

2 questions · 30 marks

112 marks

A ball is dropped from a height of 3 m. After each bounce it rises to 3/4 of the height it fell from.

  1. (a)
    How high does the ball rise after the 5th 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 8th 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 15 cm?
    (3)
Show the answers

(a) Answer: 0.71 m

  1. a geometric sequence of heights with r = 3/4
  2. 3 × (3/4)^5 = 0.71 m

(b) Answer: 18.60 m

  1. down 3 m, then up and down again after each of 7 bounces
  2. 3 + 2 × 3(3/4)[1 − (3/4)^7]/(1 − 3/4)
  3. = 18.60 m

(c) Answer: 21.00 m

  1. −1 < r = 3/4 < 1, so the sum converges
  2. 3 + 2 × 3(3/4)/(1 − 3/4)
  3. = 21.00 m

(d) Answer: after bounce 11

  1. 3(3/4)^n < 0.15
  2. n > log(0.15/3)/log(3/4) = 10.41
  3. n = 11
218 marks
  1. (a)
    In a geometric sequence T_4 = −54 and T_7 = −1458. Determine the first term and the common ratio.
    (4)
  2. (b)
    Calculate the sum of the first 8 terms.
    (2)
  3. (c)
    Determine T_(8).
    (2)
  4. (d)
    Which is the first term of the sequence whose size is greater than 30000?
    (4)
  5. (e)
    For which values of x will the series Σ_(n=1)^(∞) ((x + 4)/3)^n converge?
    (3)
  6. (f)
    Calculate the sum to infinity of the series when x = −3.
    (3)
Show the answers

(a) Answer: a = −2 and r = 3

  1. ar^3 = −54 and ar^6 = −1458
  2. r^3 = −1458/−54 = 27
  3. r = 3
  4. a = −54/(3)^3 = −2

(b) Answer: S_(8) = −6560

  1. S_(8) = −2((3)^8 − 1)/(3 − 1)
  2. = −6560

(c) Answer: T_(8) = −4374

  1. T_(8) = −2(3)^7
  2. = −4374

(d) Answer: T_(10)

  1. |−2|·3^(n − 1) > 30000
  2. n − 1 > log(30000/2)/log 3
  3. n = 10

(e) Answer: −7 < x < −1

  1. a geometric series with r = (x + 4)/3
  2. −1 < (x + 4)/3 < 1
  3. −7 < x < −1

(f) Answer: S_(∞) = 1/2

  1. r = (−3 + 4)/3 = 1/3
  2. a = T_1 = 1/3
  3. S_(∞) = a/(1 − r) = 1/2

Arithmetic sequences

2 questions · 34 marks

317 marks
  1. (a)
    2x + 1; 4x + 1 and x + 11 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 97?
    (3)
  3. (c)
    How many terms of this sequence must be added to give a sum of 434?
    (4)
  4. (d)
    A pattern of dots is built picture by picture. Pictures 1 to 4 have 7; 19; 37; 61 dots, and the numbers of dots form a quadratic sequence. How many dots are in picture 20?
    (4)
  5. (e)
    Which picture has 2269 dots?
    (3)
Show the answers

(a) Answer: x = 2; the terms are 5; 9; 13

  1. T_2 − T_1 = T_3 − T_2
  2. 2(4x + 1) = (2x + 1) + (x + 11)
  3. 5x − 10 = 0
  4. x = 2

(b) Answer: T_(24)

  1. 5 + (n − 1)(4) = 97
  2. n − 1 = 23
  3. n = 24

(c) Answer: 14 terms

  1. n/2[2(5) + (n − 1)(4)] = 434
  2. 4n^2 + 6n − 868 = 0
  3. n = 14 (n is a natural number)

(d) Answer: 1261 dots

  1. second difference 6: a = 3
  2. T_n = 3n^2 + 3n + 1
  3. T_(20) = 1261

(e) Answer: picture 27

  1. 3n^2 + 3n + 1 = 2269
  2. 3n^2 + 3n − 2268 = 0
  3. n = 27 (n > 0)
417 marks

Consider the arithmetic series 19 + 22 + 25 + 28 + … + 157

  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 −14; −26; −42; −62; −86; … Determine T_n, the nth term.
    (4)
  6. (f)
    Determine whether −762 is a term of this quadratic sequence.
    (3)
Show the answers

(a) Answer: T_n = 3n + 16

  1. a = 19 and d = 22 − (19) = 3
  2. T_n = 19 + (n − 1)(3) = 3n + 16

(b) Answer: 47 terms

  1. 3n + 16 = 157
  2. 3n = 141
  3. n = 47

(c) Answer: S_(47) = 4136

  1. S_n = n/2(a + l)
  2. S_(47) = 47/2(19 + 157)
  3. = 4136

(d) Answer: Σ_(n=1)^(47) (3n + 16)

  1. the terms are T_n = 3n + 16, from n = 1 to n = 47
  2. Σ_(n=1)^(47) (3n + 16)

(e) Answer: T_n = −2n^2 − 6n − 6

  1. first differences: −12; −16; −20; −24
  2. second difference: −4, so 2a = −4 and a = −2
  3. 3a + b = −12, so b = −6
  4. a + b + c = −14, so c = −6
  5. T_n = −2n^2 − 6n − 6

(f) Answer: Yes: it is T_(18)

  1. −2n^2 − 6n − 6 = −762
  2. −2n^2 − 6n + 756 = 0
  3. n = 18, a natural number

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