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Admissions tests / TMUA / Paper 1 / Sequences and series

Test standard. 12 questions, 12 marks, about 48 minutes.

TMUA Paper 1: Sequences and series, set 2

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

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

    A sequence is defined by x_1 = 2 and x_(n+1) = 1/(1 - x_n) for n >= 1. Find x_100.

    1. A 2
    2. B -1
    3. C 1/2
    4. D -295
  2. 21 mark

    A sequence is defined by x_1 = 4 and x_(n+1) = (x_n + 6)/2 for n >= 1. As n becomes large, x_n approaches a limiting value L. Find L.

    1. A 5
    2. B 6
    3. C 4
    4. D 2
  3. 31 mark

    A sequence has nth term u_n = n^2 - n + 1 for positive integers n. Find the positive integer n for which u_n = 91.

    1. A 19
    2. B -9
    3. C 10
    4. D 9
  4. 41 mark

    The sum of the first n terms of an arithmetic series is given by S_n = 3n^2 + 2n. Find the 10th term of the series.

    1. A 53
    2. B 32
    3. C 320
    4. D 59
  5. 51 mark

    An arithmetic series has first term 6. The sum of the first 9 terms of the series equals the sum of the first 4 terms. Find the common difference.

    1. A 1
    2. B 6
    3. C -1/2
    4. D -1
  6. 61 mark

    Find the sum of all integers from 1 to 200 inclusive that are NOT multiples of 5.

    1. A 20100
    2. B 16200
    3. C 16000
    4. D 20060
  7. 71 mark

    A geometric series has all terms positive. Its second term is 6 and its fourth term is 54. Find the sum of the first 4 terms of the series.

    1. A 26
    2. B 240
    3. C 40
    4. D 80
  8. 81 mark

    A ball is dropped from a height of 4 metres. Each time it hits the ground it bounces back up to 2/3 of the height it fell from, and this continues forever. Find the total distance travelled by the ball.

    1. A 20
    2. B 12
    3. C 28
    4. D 16
  9. 91 mark

    A geometric series converges and its sum to infinity is three times its first term. Find the common ratio r.

    1. A -2
    2. B 2/3
    3. C 3
    4. D -2/3
  10. 101 mark

    Find the coefficient of x^4 in the expansion of (3 - x^2)^5.

    1. A 90
    2. B -270
    3. C 270
    4. D 15
  11. 111 mark

    Find the coefficient of x^3 in the expansion of (1 - 2x)^5.

    1. A 40
    2. B -120
    3. C -80
    4. D 80
  12. 121 mark

    Given that C(n,3) = 4 x C(n,2), find the positive integer n.

    1. A 6
    2. B 14
    3. C 10
    4. D 12

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. Compute the first few terms directly: x_2 = 1/(1-2) = -1, x_3 = 1/(1-(-1)) = 1/2, and x_4 = 1/(1 - 1/2) = 2.
    2. Since x_4 = x_1, the sequence repeats every 3 terms from here on: the values cycle 2, -1, 1/2, 2, -1, 1/2, ...
    3. Since 99 = 3 x 33, term 100 sits exactly 99 steps (a whole number of complete 3-term cycles) after term 1, so x_100 occupies the same position in the cycle as x_1.
    4. Therefore x_100 = x_1 = 2.
    • Why not B: Comes from matching the remainder when 100 is divided by 3 to the wrong position in the repeating block, taking a remainder of 1 as the position of x_2 rather than x_1, and so reading off x_2's value of -1.
    • Why not C: Comes from miscalculating 100 mod 3 as 0 rather than 1 (for instance by wrongly treating 100 as a multiple of 3), which matches to x_3's value of 1/2 instead of x_1's.
    • Why not D: Comes from wrongly treating the recurrence as if it were an arithmetic sequence with common difference d = x_2 - x_1 = -3, and applying x_n = x_1 + (n-1)d = 2 + 99(-3) = -295, without ever noticing that the sequence is not arithmetic at all and in fact repeats every 3 terms.

  2. Question 2Answer: B

    1. As n becomes large, x_n and x_(n+1) both approach the same limiting value L, so L must satisfy the recurrence with both x_n and x_(n+1) replaced by L: L = (L+6)/2.
    2. Multiplying both sides by 2 gives 2L = L + 6.
    3. Subtracting L from both sides gives L = 6.
    4. This is consistent with the terms themselves: 4, 5, 5.5, 5.75, 5.875, ..., which climb steadily towards 6 without ever reaching or passing it.
    • Why not A: Comes from mistaking a single early iterate, x_2 = (4+6)/2 = 5, for the eventual limit, stopping after one application of the recurrence rather than solving the fixed-point equation.
    • Why not C: Comes from assuming the given first term is already the limit, since the terms look like they change slowly, rather than recognising the sequence is still climbing towards a different value.
    • Why not D: Comes from moving the L term to the wrong side of the equation 2L = L + 6 with the wrong sign, writing 2L + L = 6 instead of 2L - L = 6, which gives 3L = 6 and so L = 2.

  3. Question 3Answer: C

    1. Set u_n = 91: n^2 - n + 1 = 91, so n^2 - n - 90 = 0.
    2. This factorises as (n-10)(n+9) = 0, since -10 x 9 = -90 and -10 + 9 = -1, matching the coefficient of n.
    3. So n = 10 or n = -9. Since n must be a positive integer, n = -9 is rejected.
    4. Therefore n = 10. Check: u_10 = 100 - 10 + 1 = 91, as required.
    • Why not A: Comes from taking the square root of the discriminant, 19, to be the value of n itself, skipping the final step of the quadratic formula that requires adding 1 and dividing by 2.
    • Why not B: Comes from correctly solving the quadratic but taking the negative root, n = -9, without checking the requirement that n be a positive integer.
    • Why not D: Comes from attempting to factorise n^2 - n - 90 as (n-9)(n+10) instead of the correct (n-10)(n+9), swapping which bracket carries the minus sign, and taking the resulting root n = 9.

  4. Question 4Answer: D

    1. The nth term of a series is related to its partial sums by u_n = S_n - S_(n-1).
    2. S_10 = 3(10)^2 + 2(10) = 300 + 20 = 320.
    3. S_9 = 3(9)^2 + 2(9) = 243 + 18 = 261.
    4. u_10 = S_10 - S_9 = 320 - 261 = 59.
    • Why not A: Comes from computing u_9 = S_9 - S_8 instead of u_10 = S_10 - S_9, an off-by-one error in which two consecutive partial sums are subtracted.
    • Why not B: Comes from assuming S_n = n x u_n, as if every term equalled the mean term, and dividing S_10 = 320 by 10 to get 32, which only holds for a constant sequence.
    • Why not C: Comes from mistaking the partial sum S_10 itself for the 10th term u_10, forgetting that u_10 = S_10 - S_9 rather than S_10 alone.

  5. Question 5Answer: D

    1. Using S_n = (n/2)[2a + (n-1)d] with a = 6: S_9 = (9/2)[12 + 8d] = 54 + 36d, and S_4 = (4/2)[12 + 3d] = 24 + 6d.
    2. Setting S_9 = S_4: 54 + 36d = 24 + 6d.
    3. Rearranging: 30d = -30, so d = -1.
    4. This makes sense: with a negative common difference, the extra terms between the 5th and 9th positions are negative enough to exactly cancel the surplus, bringing S_9 back down to equal S_4.
    • Why not A: Comes from a sign slip in the final division, treating 30d = -30 as giving d = 1 rather than d = -1.
    • Why not B: Comes from forgetting to multiply through by n/2 when forming S_9, using 12+8d directly as if it were the whole sum instead of (9/2)(12+8d), and only then setting it equal to the correctly formed S_4 = 24+6d.
    • Why not C: Comes from using a instead of 2a in the bracket 2a+(n-1)d for both sums (forgetting to double the first term), giving 27+36d = 12+6d and so d = -1/2.

  6. Question 6Answer: C

    1. The sum of all integers from 1 to 200 is (200)(201)/2 = 20100.
    2. The multiples of 5 from 5 to 200 are 5, 10, ..., 200, which is 200/5 = 40 terms. Their sum is 5(1+2+...+40) = 5 x (40)(41)/2 = 5 x 820 = 4100.
    3. The sum of the integers that are NOT multiples of 5 is the total sum minus the sum of the multiples of 5: 20100 - 4100 = 16000.
    • Why not A: Comes from computing the sum of all integers from 1 to 200 but forgetting to exclude the multiples of 5 at all, leaving the full total unchanged.
    • Why not B: Comes from undercounting the multiples of 5 up to 200 as 39 instead of 40 (forgetting that 200 itself is a multiple of 5), giving a sum of multiples of 3900 instead of 4100 and so an answer that is 200 too high.
    • Why not D: Comes from subtracting the COUNT of multiples of 5 (40) rather than their SUM (4100) from the total, confusing 'how many terms to remove' with 'how much value to remove'.

  7. Question 7Answer: D

    1. Let the first term be a and common ratio r. The second term is ar=6 and the fourth term is ar^3=54.
    2. Dividing: ar^3/(ar) = r^2 = 54/6 = 9, so r = 3 or r = -3.
    3. If r=3, a=6/3=2, giving terms 2, 6, 18, 54: all positive, as required. If r=-3, a=6/(-3)=-2, giving terms -2, 6, -18, 54: not all positive, so this case is rejected.
    4. With a=2 and r=3: sum of the first 4 terms = a(r^4-1)/(r-1) = 2(81-1)/2 = 80.
    • Why not A: Comes from computing the sum of the first 3 terms instead of the first 4, an off-by-one in the number of terms used in the sum formula.
    • Why not B: Comes from mistaking the given second term for the first term a (an indexing slip along the sequence), using a=6 with r=3 instead of the correct a=2.
    • Why not C: Comes from using the algebraically valid but rejected case r=-3, a=-2 anyway, without checking that this case does not actually give a series of all positive terms as the question requires.

  8. Question 8Answer: A

    1. The ball falls 4 m, then each bounce reaches 2/3 of the height of the previous fall, and it falls back down that same height, forever.
    2. The bounce heights form a geometric sequence 4x(2/3), 4x(2/3)^2, ... with first term 8/3 and common ratio 2/3, so their sum to infinity is (8/3)/(1-2/3) = 8.
    3. Every bounce height is travelled twice, once up and once down, so the total distance from bouncing is 2 x 8 = 16.
    4. Adding the initial one-way drop of 4 m: total distance = 4 + 16 = 20.
    • Why not B: Comes from counting only the upward part of each bounce, forgetting that the ball also falls back down through the same height, so the bounce distances are never doubled.
    • Why not C: Comes from taking the first bounce height itself to be 4 (the original drop height), forgetting to multiply by the ratio 2/3 before starting the geometric series, which overstates every bounce.
    • Why not D: Comes from correctly doubling the bounce distances but forgetting to add the initial one-way drop of 4 metres before the bouncing begins.

  9. Question 9Answer: B

    1. For a convergent geometric series, the sum to infinity is S = a/(1-r).
    2. Given S = 3a, and a != 0: a/(1-r) = 3a, so dividing both sides by a gives 1/(1-r) = 3.
    3. Taking reciprocals: 1-r = 1/3.
    4. So r = 1 - 1/3 = 2/3, and |2/3| < 1 confirms this is consistent with the series converging.
    • Why not A: Comes from confusing 1/(1-r) = 3 with 1-r = 3, forgetting to take the reciprocal of both sides before solving, which gives r = 1 - 3 = -2.
    • Why not C: Comes from mistaking the given multiple (3) for the common ratio r itself, without ever forming or solving the equation a/(1-r)=3a.
    • Why not D: Comes from a sign slip isolating r from 1-r=1/3, computing r=(1/3)-1=-2/3 instead of the correct r=1-(1/3)=2/3.

  10. Question 10Answer: C

    1. The general term in the expansion of (3 - x^2)^5 is C(5,k) x 3^(5-k) x (-x^2)^k = C(5,k) x 3^(5-k) x (-1)^k x^(2k).
    2. The coefficient of x^4 needs 2k=4, so k=2.
    3. C(5,2) = 10, 3^(5-2) = 3^3 = 27, and (-1)^2 = 1, so the coefficient is 10 x 27 x 1 = 270.
    • Why not A: Comes from mixing up the two exponents in the general term C(5,k) 3^(5-k) (-x^2)^k, using 3^k instead of 3^(5-k): with k=2 this gives C(5,2) x 3^2 x 1 = 10 x 9 = 90.
    • Why not B: Comes from assuming (-x^2)^k must be negative since the base -x^2 is negative, without noticing that k=2 is even so (-1)^2 = 1: this gives 10 x 27 x (-1) = -270 instead of the correct +270.
    • Why not D: Comes from treating the power of x in the general term as k rather than 2k (forgetting that x^2 raised to the kth power contributes 2k), so solving k=4 directly instead of 2k=4: C(5,4) x 3^1 x 1 = 5 x 3 = 15.

  11. Question 11Answer: C

    1. The general term in the expansion of (1-2x)^5 is C(5,k) (1)^(5-k) (-2x)^k = C(5,k) (-2)^k x^k.
    2. The coefficient of x^3 comes from k=3: C(5,3) x (-2)^3.
    3. C(5,3) = 10 and (-2)^3 = -8, so the coefficient is 10 x (-8) = -80.
    • Why not A: Comes from confusing 'coefficient of x^3' with 'the 3rd term of the expansion': since the general term T_(k+1) corresponds to x^k, the 3rd term (k=2) is C(5,2)(-2)^2 x^2 = 10 x 4 x^2, and its coefficient, 40, is read off instead of using k=3.
    • Why not B: Comes from miscalculating the binomial coefficient as C(5,3)=15 instead of the correct value 10, giving 15 x (-8) = -120.
    • Why not D: Comes from dropping the sign on the negative base, computing (-2)^3 as if it were 2^3=8 instead of -8, giving 10 x 8 = 80.

  12. Question 12Answer: B

    1. C(n,3) = n!/(3!(n-3)!) and C(n,2) = n!/(2!(n-2)!), so C(n,3)/C(n,2) = [2!(n-2)!]/[3!(n-3)!] = (n-2)/3.
    2. Given C(n,3) = 4 x C(n,2), the ratio C(n,3)/C(n,2) = 4, so (n-2)/3 = 4.
    3. Solving: n-2=12, so n=14.
    4. Check: C(14,2)=91 and C(14,3)=364, and 364 = 4 x 91, confirming n=14.
    • Why not A: Comes from dropping the /3 factor in the ratio C(n,3)/C(n,2) = (n-2)/3, treating it as simply (n-2), so solving n-2=4 gives n=6.
    • Why not C: Comes from using the wrong factorial ratio (n-2)/2 instead of the correct (n-2)/3, confusing which factorial (2! versus 3!) supplies the denominator, so solving (n-2)/2=4 gives n-2=8 and n=10.
    • Why not D: Comes from correctly reaching n-2=12 but then forgetting to add the 2 back at the end, giving n=12 instead of n=14.

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