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ESAT Maths 2 Sequences & series questions

9 original ESAT Maths 2 sequences & series questions in the real exam style, each with a step-by-step worked solution and an explanation of why the tempting wrong answers are wrong. Try each one before opening the solution.

Methods on this page: Find terms from the sum formula: uₙ = Sₙ − Sₙ₋₁Set up equations from sums to infinitySpot a repeating sequence

Question 1ESAT Maths 2Warm-up

The sum of the first nn terms of an arithmetic sequence is Sn=2n2+3nS_n = 2n^2 + 3n. What is the 6th term?
  1. A1515
  2. B2424
  3. C2525
  4. D2626
  5. E2929
  6. F9090
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Answer: C · Method: Find terms from the sum formula: uₙ = Sₙ − Sₙ₋₁

  1. The 6th term is the sum of the first 6 terms minus the sum of the first 5: u6=S6−S5u_{6} = S_{6} - S_{5}.
  2. S6=90S_{6} = 90 and S5=65S_{5} = 65, so u6=25u_{6} = 25.

Why the tempting wrong answers are wrong

  • E: That's S7−S6S_{7} - S_{6}, the 7th term.
  • F: That's S6S_{6}, the sum of the first 6 terms, not the 6th term.

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Question 2ESAT Maths 2Standard

An infinite geometric series has sum 1818. The series formed by squaring each of its terms has sum 3245\dfrac{324}{5}. What is the first term of the original series?
  1. A185\dfrac{18}{5}
  2. B55
  3. C66
  4. D77
  5. E725\dfrac{72}{5}
  6. F1818
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Answer: C · Method: Set up equations from sums to infinity

  1. With first term aa and ratio rr: a1−r=18\dfrac{a}{1-r} = 18 and a21−r2=3245\dfrac{a^2}{1-r^2} = \dfrac{324}{5}.
  2. Since 1−r2=(1−r)(1+r)1 - r^2 = (1-r)(1+r), dividing the second by the first gives a1+r=185\dfrac{a}{1+r} = \dfrac{18}{5}.
  3. Solving the two linear equations gives r=23r = \dfrac{2}{3} and a=6a = 6.

Why the tempting wrong answers are wrong

  • A: That's a1+r\tfrac{a}{1+r}, not aa. Use it with a1−r\tfrac{a}{1-r} to find rr first.

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Question 3ESAT Maths 2Standard

An infinite geometric series has sum 103\dfrac{10}{3}. The series formed by squaring each of its terms has sum 1003\dfrac{100}{3}. What is its common ratio?
  1. A−52-\dfrac{5}{2}
  2. B−32-\dfrac{3}{2}
  3. C−12-\dfrac{1}{2}
  4. D14\dfrac{1}{4}
  5. E12\dfrac{1}{2}
  6. F32\dfrac{3}{2}
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Answer: C · Method: Set up equations from sums to infinity

  1. With first term aa and ratio rr: a1−r=103\dfrac{a}{1-r} = \dfrac{10}{3} and a21−r2=1003\dfrac{a^2}{1-r^2} = \dfrac{100}{3}.
  2. Since 1−r2=(1−r)(1+r)1 - r^2 = (1-r)(1+r), dividing the second by the first gives a1+r=10\dfrac{a}{1+r} = 10.
  3. Solving the two linear equations gives r=−12r = -\dfrac{1}{2} and a=5a = 5.

Why the tempting wrong answers are wrong

  • E: Check the sign of the ratio.

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Question 4ESAT Maths 2Standard

A sequence is defined by x1=−2x_1 = -2 and xn+1=xn−1xn+1x_{n+1} = \dfrac{x_n - 1}{x_n + 1} for n≥1n \ge 1. What is ∑n=130xn\displaystyle\sum_{n=1}^{30} x_n?
  1. A376\dfrac{37}{6}
  2. B436\dfrac{43}{6}
  3. C496\dfrac{49}{6}
  4. D556\dfrac{55}{6}
  5. E293\dfrac{29}{3}
  6. F616\dfrac{61}{6}
  7. G676\dfrac{67}{6}
  8. H736\dfrac{73}{6}
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Answer: D · Method: Spot a repeating sequence

  1. Work out the first few terms: −2, 3, 12, −13, −2-2,\ 3,\ \dfrac{1}{2},\ -\dfrac{1}{3},\ -2. The sequence repeats every 4 terms.
  2. Each block of 4 sums to 76\dfrac{7}{6}. There are 7 complete blocks in the first 30 terms, giving 496\dfrac{49}{6}.
  3. The remaining 2 terms are −2, 3-2,\ 3, so the total is 556\dfrac{55}{6}.

Why the tempting wrong answers are wrong

  • A: That's the sum of the first 29 terms.
  • C: That's 7 complete cycles: the last 2 terms still need adding.

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Question 5ESAT Maths 2Standard

A sequence is defined by x1=−3x_1 = -3 and xn+1=11−xnx_{n+1} = \dfrac{1}{1 - x_n} for n≥1n \ge 1. What is ∑n=125xn\displaystyle\sum_{n=1}^{25} x_n?
  1. A−493-\dfrac{49}{3}
  2. B−463-\dfrac{46}{3}
  3. C−433-\dfrac{43}{3}
  4. D−16912-\dfrac{169}{12}
  5. E−403-\dfrac{40}{3}
  6. F−373-\dfrac{37}{3}
  7. G−343-\dfrac{34}{3}
  8. H−313-\dfrac{31}{3}
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Answer: C · Method: Spot a repeating sequence

  1. Work out the first few terms: −3, 14, 43, −3-3,\ \dfrac{1}{4},\ \dfrac{4}{3},\ -3. The sequence repeats every 3 terms.
  2. Each block of 3 sums to −1712-\dfrac{17}{12}. There are 8 complete blocks in the first 25 terms, giving −343-\dfrac{34}{3}.
  3. The remaining 1 term is −3-3, so the total is −433-\dfrac{43}{3}.

Why the tempting wrong answers are wrong

  • G: That's the sum of the first 24 terms.

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Question 6ESAT Maths 2Standard

A sequence is defined by x1=2x_1 = 2 and xn+1=1−1xnx_{n+1} = 1 - \dfrac{1}{x_n} for n≥1n \ge 1. What is ∑n=125xn\displaystyle\sum_{n=1}^{25} x_n?
  1. A1212
  2. B1313
  3. C1414
  4. D292\dfrac{29}{2}
  5. E1515
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Answer: C · Method: Spot a repeating sequence

  1. Work out the first few terms: 2, 12, −1, 22,\ \dfrac{1}{2},\ -1,\ 2. The sequence repeats every 3 terms.
  2. Each block of 3 sums to 32\dfrac{3}{2}. There are 8 complete blocks in the first 25 terms, giving 1212.
  3. The remaining 1 term is 22, so the total is 1414.

Why the tempting wrong answers are wrong

  • A: That's the sum of the first 24 terms.

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Question 7ESAT Maths 2Standard

An arithmetic progression has first term aa and common difference dd. The sum of the first 10 terms plus the sum of the first 11 terms is equal to the sum of the first 15 terms. Which of the following is a correct expression for aa in terms of dd?
  1. Aa=−56da = -\dfrac{5}{6}d
  2. Ba=−16da = -\dfrac{1}{6}d
  3. Ca=56da = \dfrac{5}{6}d
  4. Da=53da = \dfrac{5}{3}d
  5. Ea=116da = \dfrac{11}{6}d
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Answer: C · Method: Find terms from the sum formula: uₙ = Sₙ − Sₙ₋₁

  1. Sn=n2(2a+(n−1)d)=na+n(n−1)2dS_n = \tfrac{n}{2}\big(2a + (n-1)d\big) = na + \tfrac{n(n-1)}{2}d.
  2. So 21a+100d=15a+105d21a + 100d = 15a + 105d.
  3. Rearranging: a=56da = \dfrac{5}{6}d.

Why the tempting wrong answers are wrong

  • A: Check the sign when you rearrange for aa.
  • B: Sn=n2(2a+(n−1)d)S_n = \tfrac{n}{2}\big(2a + (n - 1)d\big): it's n−1n - 1, not n+1n + 1.

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Question 8ESAT Maths 2Challenging

The first three terms of a convergent geometric progression are p, p−220, p−165p,\ p - 220,\ p - 165. What is the sum to infinity of the progression?
  1. A−7045-\dfrac{704}{5}
  2. B6995\dfrac{699}{5}
  3. C7045\dfrac{704}{5}
  4. D7095\dfrac{709}{5}
  5. E7043\dfrac{704}{3}
  6. F14085\dfrac{1408}{5}
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Answer: C · Method: Set up equations from sums to infinity

  1. In a GP the middle term squared equals the product of its neighbours: (p−220)2=p(p−165)(p - 220)^2 = p(p - 165).
  2. Solving gives p=176p = 176, so the terms are 176,−44,11176, -44, 11, so r=−14r = -\dfrac{1}{4}.
  3. Sum to infinity =1761−(−14)=7045= \dfrac{176}{1 - (-\dfrac{1}{4})} = \dfrac{704}{5}.

Why the tempting wrong answers are wrong

  • E: The sum to infinity is a1−r\tfrac{a}{1 - r}.

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Question 9ESAT Maths 2Challenging

The first three terms of a convergent geometric progression are 3p, p−24, p+123p,\ p - 24,\ p + 12. What is the sum to infinity of the progression?
  1. A−576-576
  2. B−289-289
  3. C−288-288
  4. D−287-287
  5. E−286-286
  6. F−96-96
  7. G99
  8. H288288
Show the answer and worked solution

Answer: C · Method: Set up equations from sums to infinity

  1. In a GP the middle term squared equals the product of its neighbours: (p−24)2=3p(p+12)(p - 24)^2 = 3p(p + 12).
  2. Solving gives p=−48p = -48 or p=6p = 6. Only p=−48p = -48 gives ∣r∣<1|r| < 1: the terms are
  3. Sum to infinity =−1441−(12)=−288= \dfrac{-144}{1 - (\dfrac{1}{2})} = -288.

Why the tempting wrong answers are wrong

  • F: The sum to infinity is a1−r\tfrac{a}{1 - r}.
  • G: That root gives a common ratio with ∣r∣≥1|r| \ge 1, so that progression doesn't converge.

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