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Admissions tests / TMUA / Paper 1 / Sequences and series

Foundation. 12 questions, 12 marks, about 45 minutes.

TMUA Paper 1: Sequences and series, set 1

Sequences given by a formula or a recurrence, arithmetic and geometric series, sigma notation and the binomial expansion.

Download the questions (PDF) Download with worked solutions (PDF)

  • Answer all questions. No calculator.
  • Each question has exactly one correct answer.
  1. 11 mark

    A sequence is defined by x_1 = 5 and x_(n+1) = 2x_n - 3 for n >= 1. Find x_4.

    1. A 19
    2. B 11
    3. C 22
    4. D 61
  2. 21 mark

    A sequence has nth term u_n = n^2 - 7n + 10 for positive integers n. For how many positive integers n is u_n negative?

    1. A 2
    2. B 1
    3. C 3
    4. D 4
  3. 31 mark

    A sequence is defined by x_1 = 2 and x_(n+1) = (x_n)^2 - 1 for n >= 1. Find x_3.

    1. A 3
    2. B 8
    3. C 5
    4. D 26
  4. 41 mark

    Find the sum of the first 20 terms of an arithmetic series with first term 3 and common difference 4.

    1. A 650
    2. B 860
    3. C 1640
    4. D 820
  5. 51 mark

    Find the sum of all multiples of 3 from 1 to 100 inclusive.

    1. A 1683
    2. B 561
    3. C 1584
    4. D 1785
  6. 61 mark

    The 5th term of an arithmetic series is 17 and the 12th term is 45. Find the sum of the first 10 terms of the series.

    1. A 150
    2. B 190
    3. C 210
    4. D 380
  7. 71 mark

    A geometric series has first term 8 and common ratio 1/3. Find the sum to infinity.

    1. A 6
    2. B 8
    3. C 16/3
    4. D 12
  8. 81 mark

    A geometric series has sum to infinity 20 and first term 5. Find the common ratio r.

    1. A 1/4
    2. B -3
    3. C 3/4
    4. D 4/5
  9. 91 mark

    Find the sum of the first 6 terms of the geometric series with first term 2 and common ratio 3.

    1. A 57
    2. B 242
    3. C 364
    4. D 728
  10. 101 mark

    Find the coefficient of x^3 in the expansion of (1+x)^8.

    1. A 112
    2. B 70
    3. C 336
    4. D 56
  11. 111 mark

    Find the coefficient of x^2 in the expansion of (2+3x)^4.

    1. A 72
    2. B 216
    3. C 108
    4. D 288
  12. 121 mark

    Given that C(n,2) = 45, find the positive integer n.

    1. A 9
    2. B 45
    3. C 10
    4. D 90

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

  1. Question 1Answer: A

    1. The recurrence x_(n+1) = 2x_n - 3 with x_1 = 5 means each new term comes from applying this rule once to the previous term.
    2. x_2 = 2(5) - 3 = 10 - 3 = 7.
    3. x_3 = 2(7) - 3 = 14 - 3 = 11.
    4. x_4 = 2(11) - 3 = 22 - 3 = 19, reached after three applications of the rule starting from x_1.
    • Why not B: Comes from applying the recurrence only twice, finding x_2 and x_3, and then mislabelling x_3 = 11 as x_4: an off-by-one indexing error.
    • Why not C: Comes from correctly reaching x_3 = 11 but then forgetting to subtract 3 on the final step, computing 2 x 11 = 22 instead of 2 x 11 - 3.
    • Why not D: Comes from using x_(n+1) = 2x_n + 3 throughout, reversing the sign of the constant term in the recurrence, which gives 5, 13, 29, 61.

  2. Question 2Answer: A

    1. Factorise: n^2 - 7n + 10 = (n-2)(n-5).
    2. This product is negative exactly when one factor is positive and the other negative, which happens for 2 < n < 5.
    3. The positive integers strictly between 2 and 5 are n = 3 and n = 4, so there are 2 such values.
    4. Check: u_3 = (1)(-2) = -2 and u_4 = (2)(-1) = -2, both negative, confirming the count of 2.
    • Why not B: Comes from wrongly requiring both factors (n-2) and (n-5) to be negative at once for the product to be negative, which only holds for n < 2, giving the single positive integer n = 1.
    • Why not C: Comes from including n = 5, where u_5 = (5-2)(5-5) = 0, and mistakenly treating this zero value as negative, giving the count {3, 4, 5}.
    • Why not D: Comes from including both roots n = 2 and n = 5, treating the boundary values themselves as making u_n negative rather than zero, giving the count {2, 3, 4, 5}.

  3. Question 3Answer: B

    1. Apply the recurrence once to x_1 = 2 to get x_2: x_2 = 2^2 - 1 = 4 - 1 = 3.
    2. Apply the recurrence again to x_2 = 3 to get x_3: x_3 = 3^2 - 1 = 9 - 1 = 8.
    3. So x_3 = 8, reached after two applications of the rule to the given first term.
    • Why not A: Comes from applying the recurrence only once, finding x_2 = 3, and mislabelling this as x_3: an off-by-one indexing error.
    • Why not C: Comes from confusing 'squared' with 'doubled' on the final step, computing 2 x 3 - 1 = 5 instead of 3^2 - 1.
    • Why not D: Comes from using x_(n+1) = (x_n)^2 + 1 throughout, reversing the sign of the constant term, which gives x_2 = 5 and x_3 = 26.

  4. Question 4Answer: D

    1. The sum of the first n terms of an arithmetic series is S_n = (n/2)[2a + (n-1)d], where a is the first term and d is the common difference.
    2. Here a = 3, d = 4 and n = 20, so (n-1)d = 19 x 4 = 76.
    3. 2a + (n-1)d = 6 + 76 = 82.
    4. S_20 = (20/2) x 82 = 10 x 82 = 820.
    • Why not A: Comes from swapping the roles of the first term and the common difference, using a = 4 and d = 3 instead of a = 3 and d = 4.
    • Why not B: Comes from using n in place of (n-1) in the bracket, effectively summing to a 21st term instead of the 20th: (20/2)[2(3) + 20(4)] = 10(86) = 860.
    • Why not C: Comes from forgetting to divide by 2 in the sum formula, computing n[2a + (n-1)d] instead of (n/2)[2a + (n-1)d]: 20(82) = 1640.

  5. Question 5Answer: A

    1. The multiples of 3 from 1 to 100 are 3, 6, 9, ..., 99, an arithmetic sequence with first term 3 and common difference 3.
    2. The number of terms is n = 99/3 = 33.
    3. Since each term is 3 times a natural number, the sum is 3 x (1 + 2 + ... + 33) = 3 x [33 x 34 / 2] = 3 x 561.
    4. 3 x 561 = 1683.
    • Why not B: Comes from summing the natural numbers 1 to 33 (using the sum formula) but forgetting to multiply by 3 to convert this into the sum of multiples of 3: 33 x 34 / 2 = 561.
    • Why not C: Comes from treating the sequence as starting from 0 rather than 3, effectively shifting every term down by one common difference: (33/2)[2(0) + 32(3)] = 1584.
    • Why not D: Comes from taking there to be 34 multiples of 3 up to 100 (rounding 100/3 up instead of down, so including 102): (34/2)(3 + 102) = 1785.

  6. Question 6Answer: B

    1. Let the first term be a and common difference d. The 5th term is a+4d=17 and the 12th term is a+11d=45.
    2. Subtracting gives 7d = 45 - 17 = 28, so d = 4.
    3. Then a = 17 - 4(4) = 1.
    4. S_10 = (10/2)[2a + 9d] = 5[2(1) + 9(4)] = 5[2 + 36] = 5 x 38 = 190.
    • Why not A: Comes from using the term formula a + kd for the kth term instead of a + (k-1)d, treating the 5th term as a+5d and the 12th as a+12d: this gives d = 4 but a = 17 - 20 = -3, and S_10 = 5[2(-3)+9(4)] = 150.
    • Why not C: Comes from correctly finding a = 1 and d = 4, but taking the 10th term as a + 10d = 41 instead of a + 9d = 37 before using S = (n/2)(a+l): 5(1+41) = 210.
    • Why not D: Comes from forgetting to divide by 2 in the sum formula, computing n[2a+(n-1)d] instead of (n/2)[2a+(n-1)d]: 10(38) = 380.

  7. Question 7Answer: D

    1. For a convergent geometric series with |r| < 1, the sum to infinity is S = a/(1-r).
    2. Here a = 8 and r = 1/3, so 1 - r = 1 - 1/3 = 2/3.
    3. S = 8 / (2/3) = 8 x 3/2 = 12.
    • Why not A: Comes from using S = a/(1+r) instead of S = a/(1-r), a sign error in the denominator: 8/(1+1/3) = 8/(4/3) = 6.
    • Why not B: Comes from mistaking the sum to infinity for the first term itself, never applying the formula at all.
    • Why not C: Comes from multiplying a by (1-r) instead of dividing by it: 8 x (2/3) = 16/3.

  8. Question 8Answer: C

    1. The sum to infinity formula is S = a/(1-r), so 1 - r = a/S.
    2. Here a = 5 and S = 20, so a/S = 5/20 = 1/4.
    3. Therefore 1 - r = 1/4, giving r = 1 - 1/4 = 3/4.
    4. Check: |3/4| < 1, so the series does converge, consistent with a finite sum to infinity existing.
    • Why not A: Comes from correctly finding 1-r = a/S = 1/4 but stopping there, mistaking the value of 1-r itself for r.
    • Why not B: Comes from rearranging S = a/(1-r) incorrectly, solving for (1-r) as S/a rather than a/S: 1-r = 20/5 = 4, giving r = -3.
    • Why not D: Comes from simplifying the fraction a/S = 5/20 incorrectly as 1/5 instead of 1/4, giving r = 1 - 1/5 = 4/5.

  9. Question 9Answer: D

    1. The sum of the first n terms of a geometric series is S_n = a(r^n - 1)/(r-1) for r != 1.
    2. Here a = 2, r = 3 and n = 6, so r^6 = 729.
    3. S_6 = 2(729 - 1)/(3 - 1) = 2 x 728 / 2.
    4. 2 x 728 / 2 = 728.
    • Why not A: Comes from treating the common ratio as if it were a common difference and applying the arithmetic series sum formula instead of the geometric one: (6/2)[2(2)+5(3)] = 3(19) = 57.
    • Why not B: Comes from using r^5 in place of r^6 in the formula, effectively summing only the first 5 terms: 2(3^5-1)/(3-1) = 2(242)/2 = 242.
    • Why not C: Comes from forgetting to multiply by the first term a, using (r^n-1)/(r-1) alone: (729-1)/2 = 364.

  10. Question 10Answer: D

    1. The coefficient of x^k in (1+x)^n is C(n,k) = n!/(k!(n-k)!).
    2. Here n = 8, k = 3, so the coefficient is 8!/(3! x 5!).
    3. 8!/5! = 8 x 7 x 6 = 336.
    4. Dividing by 3! = 6: 336/6 = 56.
    • Why not A: Comes from dividing by 3 instead of 3! = 6 in the combination formula, treating r! as just r: 336/3 = 112.
    • Why not B: Comes from computing C(8,4) instead of C(8,3), an off-by-one error in the lower index: C(8,4) = 70.
    • Why not C: Comes from computing 8!/5! only, which is the permutations formula, and forgetting to also divide by 3!: 8 x 7 x 6 = 336.

  11. Question 11Answer: B

    1. The general term in the expansion of (2+3x)^4 is C(4,k) x 2^(4-k) x (3x)^k.
    2. For the x^2 term, k=2, so the term is C(4,2) x 2^2 x (3x)^2.
    3. C(4,2) = 6, 2^2 = 4 and 3^2 = 9, so the coefficient is 6 x 4 x 9.
    4. 6 x 4 x 9 = 216.
    • Why not A: Comes from using 3^1 instead of 3^2 in the (3x)^2 term, forgetting to square the coefficient of x: C(4,2) x 2^2 x 3 = 6 x 4 x 3 = 72.
    • Why not C: Comes from using 2^1 instead of 2^2 for the remaining factor, forgetting to raise 2 to the power (4-2): C(4,2) x 2 x 9 = 6 x 2 x 9 = 108.
    • Why not D: Comes from using the binomial coefficient C(4,1) instead of C(4,2), miscounting which term corresponds to x^2: C(4,1) x 2^3 x 9 = 4 x 8 x 9 = 288.

  12. Question 12Answer: C

    1. C(n,2) = n(n-1)/2, so n(n-1)/2 = 45 gives n(n-1) = 90.
    2. We need two consecutive positive integers whose product is 90.
    3. 10 x 9 = 90, so n = 10 and n-1 = 9 satisfy this.
    4. Check: C(10,2) = 10!/(2! x 8!) = (10 x 9)/2 = 45, confirming n = 10.
    • Why not A: Comes from solving n(n+1)/2 = 45 instead of n(n-1)/2 = 45, a sign error in the second factor: n(n+1)=90 gives n=9.
    • Why not B: Comes from mistaking the equation C(n,2) = 45 as meaning n = 45 directly, ignoring the combination formula entirely.
    • Why not D: Comes from correctly forming n(n-1) = 90 but then reading this product off as the value of n itself, instead of solving for two consecutive integers with this product.

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