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Sequences and series

nth-term and recurrence sequences; arithmetic series; finite and infinite geometric series.

Spec MM2.1, MM2.2, MM2.3, M4.19Section 1 Part 1: AS-level pure mathematics23 questions24 real past questions
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1Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

An arithmetic series has first term \(5\) and common difference \(3\). Find the sum of the first \(40\) terms.

  1. A\(2440\)
  2. B\(2540\)
  3. C\(1270\)
  4. D\(2600\)
  5. E\(2500\)
  6. F\(2580\)
Worked solution

\(S_{40} = \frac{40}{2}\left(2\cdot5 + 39\cdot3\right) = 20(10 + 117) = 2540\).

Full method — how and why
A direct application of the arithmetic sum formula. The only decision is which version of the formula to use.
  1. Identify \(a = 5\), \(d = 3\), \(n = 40\).
    'First term' and 'common difference' name \(a\) and \(d\) exactly; \(n\) is how many terms you are adding.
  2. Use \(S_n = \frac n2\left[2a + (n-1)d\right]\): \(S_{40} = 20\left[10 + 39 \times 3\right]\).
    Use this version when you know \(a\) and \(d\); the other version \(\frac n2(a + l)\) needs the last term first.
  3. Evaluate the bracket: \(10 + 117 = 127\), so \(S_{40} = 20 \times 127 = 2540\).
    Note \((n-1)d = 39 \times 3\), not \(40 \times 3\) - there are \(39\) steps between \(40\) terms.
Where it goes wrong
  • Using \(40d\) instead of \(39d\), giving \(2600\).
  • Forgetting to halve \(n\), or halving the bracket instead.
Takeaway \(S_n = \frac n2[2a + (n-1)d]\) - the number of gaps is \(n - 1\), one fewer than the number of terms.
Answer B · Sequences and series (MM2.2) · question SEQ-01
2Sequences and series · spec MM2.3 · Paper 1 style · ★☆☆

Find the sum to infinity of the geometric series \(12 - 4 + \dfrac{4}{3} - \dfrac{4}{9} + \ldots\)

  1. A\(16\)
  2. B\(6\)
  3. C\(9\)
  4. D\(18\)
  5. E\(24\)
  6. F\(8\)
Worked solution

\(a = 12\), \(r = -\frac13\). \(S_\infty = \dfrac{12}{1 + \frac13} = \dfrac{12}{4/3} = 9\).

Full method — how and why
A sum to infinity. Find the ratio (including its sign) and check \(|r| < 1\) before using the formula.
  1. Find \(r\) by dividing consecutive terms: \(\frac{-4}{12} = -\frac13\).
    Always divide a term by the one before it; the alternating signs mean \(r\) is negative.
  2. Check convergence: \(\left|-\frac13\right| < 1\) ✓.
    The sum to infinity only exists when \(|r| < 1\); otherwise the formula is meaningless.
  3. Apply \(S_\infty = \dfrac{a}{1 - r} = \dfrac{12}{1 - (-\frac13)} = \dfrac{12}{\frac43}\).
    \(1 - r\) with a negative \(r\) becomes \(1 + \frac13\) - the double negative is the main trap.
  4. Divide: \(12 \times \frac34 = 9\).
    Dividing by a fraction is multiplying by its reciprocal.
Where it goes wrong
  • Using \(r = +\frac13\) and getting \(18\).
  • Computing \(\frac{12}{\frac43}\) as \(16\) by multiplying instead of dividing.
Takeaway \(S_\infty = \frac{a}{1-r}\); if \(r\) is negative the denominator grows, so the sum is smaller than \(a/(1-|r|)\).
Answer C · Sequences and series (MM2.3) · question SEQ-02
3Sequences and series · spec MM2.1 · Paper 1 style · ★☆☆

A sequence is defined by \(x_{n+1} = 3x_n - 2\) with \(x_1 = 2\). Find \(x_5\).

  1. A\(244\)
  2. B\(80\)
  3. C\(28\)
  4. D\(82\)
  5. E\(162\)
  6. F\(54\)
Worked solution

\(x_2 = 4\), \(x_3 = 10\), \(x_4 = 28\), \(x_5 = 82\).

Full method — how and why
A recurrence with a small index - just iterate. There is no shortcut worth finding for only four steps.
  1. Apply the rule repeatedly from \(x_1 = 2\): \(x_2 = 3(2) - 2 = 4\).
    \(x_{n+1} = 3x_n - 2\) means 'triple the last term, subtract 2'.
  2. \(x_3 = 3(4) - 2 = 10\); \(x_4 = 3(10) - 2 = 28\); \(x_5 = 3(28) - 2 = 82\).
    Write each step on its own line - the numbers grow quickly and a single slip propagates.
Where it goes wrong
  • Stopping at \(x_4 = 28\) (an offered option) by miscounting the steps: going from \(x_1\) to \(x_5\) takes four applications.
  • Reading the rule as \(3(x_n - 2)\), which gives \(x_5 = 162\) - also offered.
Takeaway Count the applications carefully: reaching \(x_k\) from \(x_1\) takes \(k - 1\) steps.
Answer D · Sequences and series (MM2.1) · question SEQ-03
4Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

The \(n\)th term of an arithmetic sequence is \(u_n = 7 - 4n\). Find the sum of the first \(30\) terms.

  1. A\(-1650\)
  2. B\(-1530\)
  3. C\(-1740\)
  4. D\(-1590\)
  5. E\(-1770\)
  6. F\(-1860\)
Worked solution

\(u_1 = 3\), \(u_{30} = 7 - 120 = -113\). \(S_{30} = \frac{30}{2}(3 - 113) = 15(-110) = -1650\).

Full method — how and why
You are given the \(n\)th term rather than \(a\) and \(d\). Generate the first and last terms and use the \(\frac n2(a + l)\) form.
  1. First term: \(u_1 = 7 - 4 = 3\). Common difference: the coefficient of \(n\) is \(-4\), so \(d = -4\).
    In \(u_n = An + B\) the gradient \(A\) is the common difference - the sequence is linear in \(n\).
  2. Last term: \(u_{30} = 7 - 120 = -113\).
    Substitute \(n = 30\) directly.
  3. Use \(S_n = \frac n2(a + l) = 15(3 + (-113)) = 15 \times (-110)\).
    This version is quickest when the first and last terms are easy to compute.
  4. Evaluate: \(-1650\).
    Negative, as expected - most terms are negative.
Where it goes wrong
  • Taking \(d = 7\) or \(a = 7\) from the formula rather than computing \(u_1\).
  • Sign slips when adding \(3 + (-113)\).
Takeaway For \(u_n = An + B\): \(d = A\) and \(u_1 = A + B\); then either sum formula works.
Answer A · Sequences and series (MM2.2) · question SEQ-04
5Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

The sum of the first \(n\) terms of a series is \(S_n = 2n^2 + 3n\). Find the \(10\)th term.

  1. A\(43\)
  2. B\(189\)
  3. C\(21\)
  4. D\(230\)
  5. E\(39\)
  6. F\(41\)
Worked solution

\(u_{10} = S_{10} - S_9 = (200 + 30) - (162 + 27) = 230 - 189 = 41\).

Full method — how and why
Given a formula for \(S_n\), individual terms come from the difference of consecutive sums.
  1. Use \(u_{10} = S_{10} - S_9\).
    The 10th term is what the total gains when you add it - this works for any series, arithmetic or not.
  2. \(S_{10} = 2(100) + 30 = 230\).
    Substitute \(n = 10\).
  3. \(S_9 = 2(81) + 27 = 189\).
    Substitute \(n = 9\); \(2 \times 81 = 162\).
  4. Subtract: \(230 - 189 = 41\).
    Both wrong intermediate values (\(230\) and \(189\)) are offered as distractors.
Where it goes wrong
  • Answering \(230\) (that is the sum of ten terms, not the tenth term).
  • Substituting \(n = 10\) into a 'term' formula that was never given.
Takeaway \(u_n = S_n - S_{n-1}\) recovers terms from a sum formula.
Answer F · Sequences and series (MM2.2) · question SEQ-06
6Sequences and series · spec MM2.3 · Paper 1 style · ★☆☆

The second term of a geometric sequence is \(6\) and the fifth term is \(48\). Find the sum of the first \(6\) terms.

  1. A\(93\)
  2. B\(186\)
  3. C\(192\)
  4. D\(381\)
  5. E\(126\)
  6. F\(189\)
Worked solution

\(ar = 6\), \(ar^4 = 48\), so \(r^3 = 8\), \(r = 2\), \(a = 3\). \(S_6 = 3\cdot\dfrac{2^6 - 1}{2 - 1} = 3 \times 63 = 189\).

Full method — how and why
Two terms of a geometric sequence determine \(a\) and \(r\). Dividing the two equations isolates \(r\).
  1. Write the facts: \(ar = 6\) and \(ar^4 = 48\).
    The \(k\)th term is \(ar^{k-1}\): the second term has \(r^1\), the fifth has \(r^4\).
  2. Divide: \(\dfrac{ar^4}{ar} = r^3 = \dfrac{48}{6} = 8\), so \(r = 2\).
    Division cancels \(a\) - always the first move with two geometric terms.
  3. Back-substitute: \(a(2) = 6\), so \(a = 3\).
    Use the simpler of the two equations.
  4. Sum: \(S_6 = a\dfrac{r^6 - 1}{r - 1} = 3 \times \dfrac{64 - 1}{1} = 189\).
    With \(r = 2\) the denominator is \(1\), making the arithmetic easy.
Where it goes wrong
  • Using \(r^4/r^1 = r^4\) instead of \(r^3\) and getting a non-integer ratio.
  • Computing \(S_6\) with \(2^6 = 32\) instead of \(64\).
Takeaway Divide two geometric terms to find \(r\); the index difference tells you which power of \(r\) you get.
Answer F · Sequences and series (MM2.3) · question SEQ-08
7Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

How many terms of the arithmetic series \(3 + 7 + 11 + \ldots\) are needed for the sum to first exceed \(500\)?

  1. A\(17\)
  2. B\(16\)
  3. C\(11\)
  4. D\(14\)
  5. E\(18\)
  6. F\(15\)
Worked solution

\(S_n = \frac n2(6 + 4(n-1)) = n(2n + 1)\). \(n = 15\): \(465\); \(n = 16\): \(528 > 500\). So \(16\).

Full method — how and why
An inequality on a sum. Form \(S_n\), then find the first integer \(n\) that works - by trial, since \(n\) must be a whole number.
  1. Identify \(a = 3\), \(d = 4\) and form the sum: \(S_n = \frac n2\left[6 + 4(n-1)\right] = \frac n2(4n + 2) = n(2n+1)\).
    Simplifying to \(n(2n+1)\) makes the trial values much easier to compute.
  2. Test \(n = 15\): \(15 \times 31 = 465\), which is less than \(500\).
    Estimate first: \(2n^2 \approx 500\) gives \(n \approx 15.8\), so start near there.
  3. Test \(n = 16\): \(16 \times 33 = 528 > 500\) ✓.
    The sum is increasing, so the first \(n\) that exceeds \(500\) is the answer.
  4. Answer: \(16\) terms.
    Check the one below to be sure it fails - that is what 'first exceeds' demands.
Where it goes wrong
  • Solving \(2n^2 + n = 500\) and rounding down to \(15\).
  • Forgetting that \(n\) must be a positive integer and quoting a decimal.
Takeaway For 'first exceeds' questions, solve approximately then test the two neighbouring integers.
Answer B · Sequences and series (MM2.2) · question SEQ-09
8Sequences and series · spec MM2.3 · Paper 1 style · ★☆☆

A geometric series has first term \(2\) and common ratio \(r > 0\). The sum of the first three terms is \(14\). Find the sum of the first five terms.

  1. A\(30\)
  2. B\(52\)
  3. C\(46\)
  4. D\(94\)
  5. E\(126\)
  6. F\(62\)
Worked solution

\(2 + 2r + 2r^2 = 14 \Rightarrow r^2 + r - 6 = 0 \Rightarrow r = 2\). \(S_5 = 2(2^5 - 1) = 62\).

Full method — how and why
The first three terms give a quadratic in \(r\); the positivity condition picks one root.
  1. Write the sum of three terms: \(2 + 2r + 2r^2 = 14\).
    Terms are \(a\), \(ar\), \(ar^2\) with \(a = 2\).
  2. Divide by \(2\) and rearrange: \(r^2 + r - 6 = 0\).
    Dividing first keeps the numbers small.
  3. Factorise: \((r+3)(r-2) = 0\), so \(r = 2\) or \(r = -3\). The condition \(r > 0\) gives \(r = 2\).
    The question's restriction exists precisely to discard the negative root.
  4. Sum five terms: \(S_5 = 2\dfrac{2^5 - 1}{2-1} = 2 \times 31 = 62\).
    \(2^5 = 32\), so the bracket is \(31\).
Where it goes wrong
  • Using \(r = -3\) and getting \(S_5 = 2 \times \frac{-243-1}{-4} = 122\).
  • Adding two more terms by hand and slipping: \(2 + 4 + 8 + 16 + 32 = 62\) is a good check.
Takeaway Conditions like '\(r > 0\)' are there to select between roots - apply them before continuing.
Answer F · Sequences and series (MM2.3) · question SEQ-10
9Sequences and series · spec MM2.3 · Paper 2 style · ★☆☆

Which of the following geometric series is/are convergent?

  1. I\(1 + 0.9 + 0.81 + \ldots\)
  2. II\(3 - 3 + 3 - 3 + \ldots\)
  3. III\(100 + 10 + 1 + 0.1 + \ldots\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

A geometric series converges if and only if \(|r| < 1\). I has \(r = 0.9\), III has \(r = 0.1\): both converge. II has \(r = -1\) and does not converge.

Full method — how and why
Convergence of a geometric series depends only on \(|r| < 1\). Compute \(r\) for each and apply the test.
  1. I: \(r = \frac{0.9}{1} = 0.9\), and \(|0.9| < 1\): converges.
    A ratio just below \(1\) still converges, though slowly.
  2. II: \(r = \frac{-3}{3} = -1\), and \(|-1| = 1\) is not less than \(1\): does not converge.
    The partial sums oscillate between \(3\) and \(0\) forever - they never settle, so there is no sum to infinity.
  3. III: \(r = \frac{10}{100} = 0.1\): converges.
    Small ratio, rapid convergence.
  4. I and III only.
    Match the verdicts to the option list.
Where it goes wrong
  • Thinking II converges to \(0\) or to \(1.5\) because the terms 'average out' - the partial sums must approach a single limit.
  • Judging by the size of the terms rather than the ratio: III starts at \(100\) but converges fastest.
Takeaway Convergence is decided by \(|r| < 1\) alone - not by the size of the first term or by whether the terms alternate.
Answer F · Sequences and series (MM2.3) · question SEQ-12
10Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

An arithmetic sequence has \(u_3 = 10\) and \(u_8 = 25\). Find the smallest value of \(n\) for which \(u_n > 100\).

  1. A\(36\)
  2. B\(32\)
  3. C\(34\)
  4. D\(31\)
  5. E\(33\)
  6. F\(35\)
Worked solution

\(5d = 15 \Rightarrow d = 3\), \(a = 4\). \(u_n = 4 + 3(n-1) = 3n + 1 > 100 \Rightarrow n > 33\), so \(n = 34\).

Full method — how and why
Two terms give \(a\) and \(d\); then solve an inequality for \(n\) and round to the correct integer.
  1. Subtract the two facts: \(u_8 - u_3 = 5d = 25 - 10 = 15\), so \(d = 3\).
    From \(u_3\) to \(u_8\) is five steps, so the difference is \(5d\) - count the gaps, not the terms.
  2. Find \(a\): \(u_3 = a + 2d = 10\), so \(a = 10 - 6 = 4\).
    The third term is \(a + 2d\), not \(a + 3d\).
  3. Write the general term: \(u_n = 4 + 3(n-1) = 3n + 1\).
    Simplify before solving - it avoids bracket errors.
  4. Solve \(3n + 1 > 100 \Rightarrow n > 33\), so the smallest integer is \(n = 34\).
    Strict inequality: \(n = 33\) gives exactly \(100\), which is not greater than \(100\).
Where it goes wrong
  • Answering \(33\), where \(u_{33} = 100\) exactly.
  • Using \(u_8 - u_3 = 8d - 3d\) correctly but then \(u_3 = a + 3d\) incorrectly.
Takeaway \(u_p\) to \(u_q\) spans \(q - p\) common differences; and 'greater than' excludes the boundary case.
Answer C · Sequences and series (MM2.2) · question SEQ-13
11Sequences and series · spec MM2.1 · Paper 1 style · ★☆☆

The sequence \(u_n = n^2 - 10n + 30\) is defined for positive integers \(n\). What is the smallest value taken by any term of the sequence?

  1. A\(21\)
  2. B\(5\)
  3. C\(6\)
  4. D\(-5\)
  5. E\(0\)
  6. F\(30\)
Worked solution

\(u_n = (n-5)^2 + 5\), minimised at \(n = 5\) with value \(5\).

Full method — how and why
A quadratic sequence. Completing the square finds the minimum - but \(n\) must be a positive integer, so check that the optimum is attainable.
  1. Complete the square: \(n^2 - 10n + 30 = (n - 5)^2 + 5\).
    Half of \(-10\) is \(-5\); \(30 - 25 = 5\).
  2. The square is smallest when \(n = 5\), giving \(u_5 = 5\).
    \((n-5)^2 \geq 0\) with equality only at \(n = 5\).
  3. \(n = 5\) is a positive integer, so the minimum is attained: the smallest term is \(5\).
    If the vertex had been at a non-integer (say \(4.5\)), you would have to test the integers either side.
Where it goes wrong
  • Giving \(n = 5\) (the position) instead of \(5\) (the value) - here they coincide, which hides the error.
  • Taking \(u_1 = 21\) as the smallest by assuming sequences always increase.
Takeaway Minimise a quadratic sequence by completing the square, then check the nearest integer values of \(n\).
Answer B · Sequences and series (MM2.1) · question SEQ-14
12Sequences and series · spec MM2.3 · Paper 1 style · ★☆☆

Evaluate \(\displaystyle\sum_{k=1}^{10} 3 \times 2^{k}\).

  1. A\(6141\)
  2. B\(3069\)
  3. C\(3072\)
  4. D\(3066\)
  5. E\(6144\)
  6. F\(6138\)
Worked solution

This is \(6 + 12 + \ldots + 6\cdot2^{9}\): first term \(6\), ratio \(2\), \(10\) terms: \(6(2^{10} - 1) = 6 \times 1023 = 6138\).

Full method — how and why
Sigma notation hiding a geometric series. Write out the first term and the ratio, and count the terms carefully.
  1. Write out the start: \(k = 1\) gives \(3 \times 2 = 6\); \(k = 2\) gives \(12\). So \(a = 6\), \(r = 2\).
    The first term is the value at the lower limit, not \(3\).
  2. Count the terms: \(k\) runs from \(1\) to \(10\), which is \(10\) terms.
    Upper minus lower plus one.
  3. Apply the sum formula: \(S_{10} = 6\dfrac{2^{10} - 1}{2 - 1} = 6 \times 1023\).
    \(2^{10} = 1024\) is worth knowing by heart.
  4. Evaluate: \(6138\).
    \(6 \times 1000 = 6000\) plus \(6 \times 23 = 138\).
Where it goes wrong
  • Taking \(a = 3\) (ignoring the \(k = 1\) factor of \(2\)), giving \(3069\).
  • Using \(2^{10}\) instead of \(2^{10} - 1\), giving \(6144\).
Takeaway In \(\sum a r^k\) the first term is the value at the lower limit; count terms as (upper \(-\) lower \(+ 1\)).
Answer F · Sequences and series (MM2.3) · question SEQ-15
13Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

The sum of the first \(n\) terms of an arithmetic series is \(S_n\). Given that \(S_{10} = 145\) and \(S_{20} = 490\), find the common difference.

  1. A\(2\)
  2. B\(1\)
  3. C\(5\)
  4. D\(4\)
  5. E\(3\)
  6. F\(\dfrac{3}{2}\)
Worked solution

\(S_{10} = 5(2a + 9d) = 145 \Rightarrow 2a + 9d = 29\); \(S_{20} = 10(2a + 19d) = 490 \Rightarrow 2a + 19d = 49\). Subtract: \(10d = 20\), \(d = 2\).

Full method — how and why
Two sums give two equations in \(a\) and \(d\). Subtracting eliminates \(a\) - and only \(d\) is wanted.
  1. \(S_{10} = \frac{10}{2}\left[2a + 9d\right] = 145\), so \(2a + 9d = 29\).
    Divide out the \(\frac n2\) factor straight away to keep the equations clean.
  2. \(S_{20} = \frac{20}{2}\left[2a + 19d\right] = 490\), so \(2a + 19d = 49\).
    Same formula with \(n = 20\).
  3. Subtract: \(10d = 20\), so \(d = 2\).
    The \(2a\) terms are identical in both equations, so subtraction removes them at once.
Where it goes wrong
  • Using \((n-1)d\) as \(10d\) and \(20d\) rather than \(9d\) and \(19d\).
  • Dividing \(490\) by \(20\) instead of by \(10\).
Takeaway Two sums, two unknowns: write both in the \(2a + (n-1)d\) form and subtract to eliminate \(a\).
Answer A · Sequences and series (MM2.2) · question SEQ-16
14Sequences and series · spec MM2.1 · Paper 1 style · ★☆☆

A sequence is defined by \(u_1 = 1\) and \(u_{n+1} = u_n + 2n + 1\). Which of the following is an expression for \(u_n\)?

  1. A\(n^2 - n + 1\)
  2. B\(2n - 1\)
  3. C\(2^{n} - 1\)
  4. D\(n^2 + 1\)
  5. E\(n(n+1)/2\)
  6. F\(n^2\)
Worked solution

\(u_2 = 4\), \(u_3 = 9\), \(u_4 = 16\): \(u_n = n^2\). Check: \((n+1)^2 = n^2 + 2n + 1\).

Full method — how and why
A recurrence that generates a familiar sequence. Compute terms, recognise the pattern, then verify it satisfies the rule.
  1. Generate terms: \(u_1 = 1\), \(u_2 = 1 + 3 = 4\), \(u_3 = 4 + 5 = 9\), \(u_4 = 9 + 7 = 16\).
    The increments \(3, 5, 7\) are the odd numbers - a strong hint.
  2. Recognise \(1, 4, 9, 16\) as the squares: conjecture \(u_n = n^2\).
    Pattern-spotting is legitimate here because the options let you verify.
  3. Verify against the recurrence: if \(u_n = n^2\) then \(u_n + 2n + 1 = n^2 + 2n + 1 = (n+1)^2 = u_{n+1}\) ✓.
    This is the crucial step: checking the formula satisfies the rule turns a guess into a proof.
Where it goes wrong
  • Choosing \(2n - 1\) (the increments) instead of the terms themselves.
  • Testing only \(u_1\), which several options match.
Takeaway Spot the pattern from a few terms, then prove it by substituting into the recurrence.
Answer F · Sequences and series (MM2.1) · question SEQ-18
15Sequences and series · spec MM2.3 · Paper 2 style · ★☆☆

A geometric series has first term \(a \neq 0\) and common ratio \(r\). \(P\): '\(-1 < r < 1\)'. \(Q\): 'the series has a sum to infinity'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Worked solution

A geometric series converges if and only if \(|r| < 1\), which is exactly \(-1 < r < 1\).

Full method — how and why
A definition-driven case: the convergence condition for a geometric series is exactly the stated inequality, so both directions hold.
  1. A geometric series with \(a \neq 0\) has a sum to infinity if and only if \(|r| < 1\).
    The partial sums \(\frac{a(1-r^n)}{1-r}\) converge precisely when \(r^n \to 0\).
  2. \(|r| < 1\) is the same statement as \(-1 < r < 1\).
    Modulus inequalities unfold into a double inequality.
  3. So \(P\) and \(Q\) are equivalent: necessary and sufficient.
    An 'if and only if' condition gives both directions at once.
Where it goes wrong
  • Forgetting the case \(r = \pm1\): \(r=1\) gives partial sums \(na\), which diverge; \(r=-1\) oscillates. Both are correctly excluded.
  • Answering 'sufficient but not necessary' from habit - here no counterexample exists.
  • Overlooking why \(a \neq 0\) is stated: with \(a = 0\) every \(r\) would give a (trivial) sum.
Takeaway Geometric series converge \(\iff |r|<1\), and then \(S_\infty = \frac{a}{1-r}\). An 'if and only if' condition is necessary and sufficient.
Answer A · Sequences and series (MM2.3) · question X2-04
16Sequences and series · spec MM2.2 · Paper 2 style · ★☆☆

An arithmetic sequence has first term \(a\) and common difference \(d\), both non-zero. Which of the following statements is/are true?

  1. IIf \(d > 0\) then every term after some point is positive.
  2. IIThe sum of the first \(n\) terms is a quadratic function of \(n\).
  3. IIIIf \(a\) and \(d\) are integers then \(S_n\) is an integer for every \(n\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I: terms \(a + (n-1)d\) eventually exceed any bound. II: \(S_n = \frac n2(2a + (n-1)d)\) is quadratic in \(n\). III: \(S_n\) is a sum of integers.

Full method — how and why
Three claims about an arithmetic sequence, settled by the \(n\)th-term and sum formulas.
  1. I: the terms are \(a+(n-1)d\) with \(d>0\), which increases without bound, so beyond some point every term is positive. True.
    A positive common difference eventually overcomes any negative start.
  2. II: \(S_n = \frac n2\left(2a+(n-1)d\right) = \frac d2 n^2 + \left(a - \frac d2\right)n\) - a quadratic in \(n\) (with \(d \neq 0\), so the \(n^2\) term survives). True.
    Expand and collect powers of \(n\); the hypothesis \(d \neq 0\) is what makes it genuinely quadratic.
  3. III: \(S_n\) is a sum of \(n\) integers \(a, a+d, \ldots\), hence an integer. True.
    Sums of integers are integers - no need to worry about the \(\frac n2\) in the formula, since the bracket is even when \(n\) is odd and vice versa.
  4. So all three.
    Match to the list.
Where it goes wrong
  • Rejecting III because the formula contains \(\frac n2\) - the product still comes out whole.
  • Rejecting II if \(d\) were \(0\) (linear) - but the question states \(d \neq 0\).
  • Reading I as 'every term is positive' rather than 'every term after some point'.
Takeaway \(S_n\) for an arithmetic series is quadratic in \(n\); check the stated conditions (\(a, d\) non-zero) before dismissing a claim.
Answer H · Sequences and series (MM2.2) · question X2-23
17Sequences and series · spec MM2.3 · Paper 1 style · ★★☆

A geometric series has first term \(a\) and common ratio \(r\) with \(|r| < 1\). Its sum to infinity is \(8\) and the sum to infinity of the squares of its terms is \(\dfrac{64}{3}\). Find \(r\).

  1. A\(-\dfrac12\)
  2. B\(\dfrac{1}{2}\)
  3. C\(\dfrac{1}{4}\)
  4. D\(\dfrac{2}{3}\)
  5. E\(\dfrac{3}{4}\)
  6. F\(\dfrac{1}{3}\)
Worked solution

\(\dfrac{a}{1-r} = 8\) and \(\dfrac{a^2}{1 - r^2} = \dfrac{64}{3}\). Dividing the second by the square of the first: \(\dfrac{1-r}{1+r} = \dfrac{64/3}{64} = \dfrac13\), so \(3 - 3r = 1 + r\) and \(r = \frac12\).

Full method — how and why
Two infinite sums, two unknowns. Dividing one equation by the (square of the) other eliminates \(a\) elegantly.
  1. Write the two facts: \(\dfrac{a}{1 - r} = 8\) and, for the squares (first term \(a^2\), ratio \(r^2\)), \(\dfrac{a^2}{1 - r^2} = \dfrac{64}{3}\).
    Squaring every term of a geometric series gives another geometric series with first term \(a^2\) and ratio \(r^2\).
  2. Square the first: \(\dfrac{a^2}{(1-r)^2} = 64\).
    Squaring makes the numerators match, so they can be eliminated by division.
  3. Divide the second equation by this: \(\dfrac{(1-r)^2}{1 - r^2} = \dfrac{64/3}{64} = \dfrac13\).
    Dividing kills \(a^2\) entirely - the standard move when two equations share a factor.
  4. Simplify using \(1 - r^2 = (1-r)(1+r)\): \(\dfrac{1 - r}{1 + r} = \dfrac13\).
    One factor of \((1-r)\) cancels.
  5. Cross-multiply: \(3 - 3r = 1 + r \Rightarrow 4r = 2 \Rightarrow r = \frac12\).
    Check: \(a = 4\), and \(\frac{16}{1 - 1/4} = \frac{64}{3}\) ✓.
Where it goes wrong
  • Using ratio \(r\) (not \(r^2\)) for the series of squares.
  • Forgetting to square the first equation before dividing, which leaves an \(a\) behind.
Takeaway Squaring a geometric series squares both \(a\) and \(r\); divide equations to eliminate the unknown you do not want.
Answer B · Sequences and series (MM2.3) · question SEQ-05
18Sequences and series · spec MM2.1 · Paper 1 style · ★★☆

A sequence satisfies \(x_{n+1} = \dfrac{1}{1 - x_n}\) with \(x_1 = 2\). Find \(x_{100}\).

  1. A\(1\)
  2. B\(2\)
  3. C\(\dfrac{1}{2}\)
  4. D\(-\dfrac12\)
  5. E\(0\)
  6. F\(-1\)
Worked solution

\(x_2 = \frac{1}{1-2} = -1\), \(x_3 = \frac{1}{1-(-1)} = \frac12\), \(x_4 = \frac{1}{1 - 1/2} = 2 = x_1\): the sequence has period \(3\). Since \(100 = 3\times33 + 1\), \(x_{100} = x_1 = 2\).

Full method — how and why
A recurrence with a huge index. Compute a few terms, spot the cycle, then use arithmetic on the index.
  1. Iterate: \(x_2 = \frac{1}{1-2} = -1\), \(x_3 = \frac{1}{1-(-1)} = \frac12\), \(x_4 = \frac{1}{1 - \frac12} = 2\).
    When an index like \(100\) appears, always compute the first few terms - the sequence is almost certainly periodic.
  2. \(x_4 = x_1\), so the sequence repeats with period \(3\): \(2, -1, \frac12, 2, -1, \frac12, \ldots\)
    Once a value repeats, the whole sequence from that point repeats identically.
  3. Reduce the index: \(100 = 3 \times 33 + 1\), so \(x_{100}\) is in the same position of the cycle as \(x_1\).
    Terms with indices differing by a multiple of the period are equal; compare remainders on division by \(3\).
  4. Therefore \(x_{100} = x_1 = 2\).
    Check the convention: the indices \(1, 4, 7, \ldots\) all leave remainder \(1\) on division by \(3\), and \(100\) does too.
Where it goes wrong
  • Finding that \(100\) leaves remainder \(1\) but then reading off \(x_0\) or \(x_3\) - line the remainders up with the actual starting index.
  • Giving up after two terms and guessing \(-1\) or \(\frac12\).
Takeaway Large-index recurrence questions are periodicity questions: find the cycle length, then work with the index modulo it.
Answer B · Sequences and series (MM2.1) · question SEQ-07
19Sequences and series · spec MM2.2 · Paper 1 style · ★★☆

The sum of the first \(n\) natural numbers is \(S\). The sum of the first \(2n\) natural numbers is \(kS\). Which of the following is true?

  1. A\(k = \dfrac{4n + 2}{n}\)
  2. B\(k = 4\) for all \(n\)
  3. C\(k = \dfrac{2(2n+1)}{n+1}\)
  4. D\(k = 2\) for all \(n\)
  5. E\(k = 4 - \dfrac{2}{n}\)
Worked solution

\(S = \frac{n(n+1)}{2}\) and the sum to \(2n\) is \(\frac{2n(2n+1)}{2} = n(2n+1)\). So \(k = \dfrac{n(2n+1)}{n(n+1)/2} = \dfrac{2(2n+1)}{n+1}\).

Full method — how and why
A ratio of two triangular numbers. The answer depends on \(n\), so the constant options can be eliminated immediately.
  1. Write both sums: \(S = \dfrac{n(n+1)}{2}\) and the sum to \(2n\) is \(\dfrac{2n(2n+1)}{2} = n(2n+1)\).
    Use \(1 + 2 + \cdots + m = \frac{m(m+1)}{2}\) with \(m = n\) and \(m = 2n\).
  2. Form the ratio: \(k = \dfrac{n(2n+1)}{\frac{n(n+1)}{2}}\).
    \(k\) is defined by 'sum to \(2n\)' \(= k \times\) 'sum to \(n\)'.
  3. Simplify: multiply top and bottom by \(2\) and cancel \(n\): \(k = \dfrac{2n(2n+1)}{n(n+1)} = \dfrac{2(2n+1)}{n+1}\).
    Cancelling \(n\) is legitimate since \(n \geq 1\).
  4. Sanity check with \(n = 1\): sums are \(1\) and \(3\), so \(k = 3\); the formula gives \(\frac{2 \times 3}{2} = 3\) ✓.
    A single numerical check rules out all the other options at once.
Where it goes wrong
  • Answering \(k = 4\) because 'doubling \(n\) roughly quadruples a quadratic' - true in the limit, false for finite \(n\).
  • Forgetting the factor \(2\) when dividing by a fraction.
Takeaway When the options are a mix of constants and formulas, substitute a small value of \(n\) - it eliminates most of them instantly.
Answer C · Sequences and series (MM2.2) · question SEQ-11
20Sequences and series · spec MM2.3 · Paper 1 style · ★★☆

A ball is dropped from a height of \(10\) m. After each bounce it rises to \(\dfrac{3}{5}\) of its previous height. Find the total distance travelled by the ball before it comes to rest.

  1. A\(30\) m
  2. B\(20\) m
  3. C\(40\) m
  4. D\(25\) m
  5. E\(50\) m
  6. F\(35\) m
Worked solution

Down: \(10 + 6 + 3.6 + \ldots = \dfrac{10}{1 - 3/5} = 25\). Up: \(6 + 3.6 + \ldots = 15\). Total \(40\) m.

Full method — how and why
A classic bouncing-ball problem: the ball goes down and up. Two geometric series, or one with a correction.
  1. Downward distances: \(10, 6, 3.6, \ldots\) - geometric with \(a = 10\), \(r = \frac35\). Sum \(= \dfrac{10}{1 - \frac35} = 25\).
    After each bounce it falls from the new height, so every rise is matched by an equal fall.
  2. Upward distances: \(6, 3.6, \ldots\) - the same series without the initial \(10\). Sum \(= 25 - 10 = 15\).
    The ball never rises \(10\) m: the first rise is \(6\) m, so the up-series starts one term later.
  3. Total: \(25 + 15 = 40\) m.
    Add the two, or use \(2 \times 25 - 10 = 40\) (double everything, then remove the un-matched first drop).
Where it goes wrong
  • Answering \(25\) by counting only the downward distances.
  • Answering \(50\) by doubling without subtracting the initial drop.
Takeaway Distance travelled counts every metre in both directions: total \(= 2S - (\text{first drop})\).
Answer C · Sequences and series (MM2.3) · question SEQ-17
21Sequences and series · spec MM2.3 · Paper 1 style · ★★☆

The first three terms of a geometric sequence are \(x\), \(x + 6\) and \(x + 15\). Find the fourth term.

  1. A\(\dfrac{81}{2}\)
  2. B\(\dfrac{125}{2}\)
  3. C\(\dfrac{75}{2}\)
  4. D\(\dfrac{225}{4}\)
  5. E\(\dfrac{125}{4}\)
  6. F\(45\)
Worked solution

Geometric: \((x+6)^2 = x(x+15) \Rightarrow x^2 + 12x + 36 = x^2 + 15x \Rightarrow x = 12\). The terms are \(12, 18, 27\) with ratio \(\frac32\), so the fourth term is \(27 \times \frac32 = \frac{81}{2}\).

Full method — how and why
Three consecutive geometric terms: the middle one squared equals the product of its neighbours. That single equation finds \(x\).
  1. Use the geometric condition \(\dfrac{x+6}{x} = \dfrac{x+15}{x+6}\), i.e. \((x+6)^2 = x(x+15)\).
    Consecutive terms have a constant ratio, so equating the two ratios gives the condition \(b^2 = ac\).
  2. Expand: \(x^2 + 12x + 36 = x^2 + 15x\).
    The \(x^2\) terms cancel, leaving a linear equation - which is why this method is quick.
  3. Solve: \(36 = 3x\), so \(x = 12\).
    The terms are \(12, 18, 27\).
  4. Find \(r = \frac{18}{12} = \frac32\) and the fourth term \(27 \times \frac32 = \frac{81}{2}\).
    The ratio need not be an integer; leave the answer as an improper fraction.
Where it goes wrong
  • Assuming the common ratio is \(\frac{x+6}{x+15}\) (upside down).
  • Stopping at the third term \(27\) or giving the ratio instead of the fourth term.
Takeaway Three terms in geometric progression satisfy \(b^2 = ac\); the \(x^2\) terms always cancel, leaving a linear equation.
Answer A · Sequences and series (MM2.3) · question SEQ-19
22Sequences and series · spec MM2.2 · Paper 1 style · ★★☆

The integers from \(1\) to \(100\) inclusive are written down. What is the sum of those that are multiples of \(3\) or \(5\) (or both)?

  1. A\(2500\)
  2. B\(2350\)
  3. C\(2733\)
  4. D\(2103\)
  5. E\(2418\)
  6. F\(2318\)
Worked solution

Multiples of \(3\): \(3 + \ldots + 99 = 3(1 + \ldots + 33) = 3 \times 561 = 1683\). Multiples of \(5\): \(5(1 + \ldots + 20) = 1050\). Multiples of \(15\): \(15(1 + \ldots + 6) = 315\). Total \(1683 + 1050 - 315 = 2418\).

Full method — how and why
A counting-and-summing problem needing inclusion-exclusion: multiples of \(15\) belong to both lists and would otherwise be counted twice.
  1. Multiples of \(3\) up to \(100\): \(3 + 6 + \cdots + 99 = 3(1 + 2 + \cdots + 33) = 3 \times \frac{33 \times 34}{2} = 3 \times 561 = 1683\).
    Factor out the \(3\) so the bracket is a plain triangular number.
  2. Multiples of \(5\): \(5(1 + \cdots + 20) = 5 \times 210 = 1050\).
    \(\lfloor 100/5 \rfloor = 20\) terms.
  3. Multiples of both, i.e. of \(15\): \(15(1 + \cdots + 6) = 15 \times 21 = 315\).
    'Or' in mathematics includes 'both', and multiples of \(15\) have been counted in each of the first two sums.
  4. Inclusion-exclusion: \(1683 + 1050 - 315 = 2418\).
    Add the two sets, subtract the overlap once.
Where it goes wrong
  • Forgetting to subtract the multiples of \(15\), giving \(2733\) - which is offered.
  • Using \(\text{lcm}(3,5) = 8\) or \(30\) instead of \(15\).
Takeaway For 'multiples of \(a\) or \(b\)', subtract the multiples of \(\text{lcm}(a,b)\) once.
Answer E · Sequences and series (MM2.2) · question SEQ-20
23Sequences and series · spec MM2.1 · Paper 2 style · ★★★

A sequence has \(u_1 = 3\) and \(u_{n+1} = \dfrac{u_n + 5}{2}\). Which of the following statements is/are true?

  1. IThe sequence is increasing.
  2. II\(u_n < 5\) for all \(n\).
  3. III\(u_n = 5 - 2^{2-n}\) for all \(n\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

\(u_2 = 4\), \(u_3 = 4.5\), \(u_4 = 4.75\). If \(u_n < 5\) then \(u_{n+1} = \frac{u_n + 5}{2} < 5\) and \(u_{n+1} - u_n = \frac{5 - u_n}{2} > 0\), so I and II hold. III: \(5 - 2^{2-n}\) gives \(3, 4, 4.5, \ldots\) and satisfies the recurrence: \(\frac{5 - 2^{2-n} + 5}{2} = 5 - 2^{1-n}\). All three are true.

Full method — how and why
A recurrence converging to a limit. Compute terms for intuition, then prove the two general claims and verify the closed form.
  1. Generate terms: \(u_1 = 3\), \(u_2 = \frac{3+5}{2} = 4\), \(u_3 = 4.5\), \(u_4 = 4.75\) - rising towards \(5\).
    The rule averages \(u_n\) with \(5\), so each term sits halfway between the previous one and \(5\).
  2. II: show \(u_n < 5\) always. If \(u_n < 5\) then \(u_{n+1} = \frac{u_n + 5}{2} < \frac{5+5}{2} = 5\); and \(u_1 = 3 < 5\). True.
    This is an induction argument in miniature: true at the start, and preserved by the rule.
  3. I: \(u_{n+1} - u_n = \frac{u_n + 5}{2} - u_n = \frac{5 - u_n}{2} > 0\) because \(u_n < 5\). So the sequence increases. True.
    Compute the difference algebraically rather than relying on the first few terms; it also shows *why* it increases - the gap to \(5\) is always positive.
  4. III: test \(5 - 2^{2-n}\): at \(n=1\) it gives \(5 - 2 = 3\) ✓, and \(\frac{(5 - 2^{2-n}) + 5}{2} = 5 - 2^{1-n}\), which is the formula at \(n+1\) ✓. True.
    Check the starting value and that the formula satisfies the recurrence - both are needed.
  5. All three are true.
    Match to the options.
Where it goes wrong
  • Concluding II from the first four terms alone - the general argument is what makes it certain.
  • Checking III only at \(n = 1\); many formulas share a first term.
Takeaway To prove a claim about every term: check the first case, then show the recurrence preserves the property.
Answer H · Sequences and series (MM2.1) · question SEQ-21

How this topic has been examined

Real TMUA questions on Sequences and series, newest first. Links open the official paper at the question and the online marking page.

2023 · Paper 1Question 18geometric series in kquestion papersit online
2023 · Paper 2Question 16recurrence with u1 = a, u2 = bquestion papersit online
2022 · Paper 1Question 5recurrence x(n+1) = x(n) + p x(n) + qquestion papersit online
2022 · Paper 1Question 8geometric series: S30 - S20 = k S10question papersit online
2021 · Paper 1Question 3arithmetic and convergent geometric progressionsquestion papersit online
2021 · Paper 2Question 10sequence 15, 21, ...question papersit online
2020 · Paper 1Question 4GP terms in an AP; sum to infinity 12question papersit online
2020 · Paper 2Question 14sum of first 4 terms = sum of first 8 termsquestion papersit online
2019 · Paper 1Question 4sequence x(n+1) = sqrt(x(n))question papersit online
2019 · Paper 1Question 5S6 = 9 S3, seventh term 360question papersit online
2019 · Paper 2Question 11integer arithmetic series summing to 20question papersit online
October 2018 · Paper 1Question 2S5 = S8 in an arithmetic progressionquestion papersit online
October 2018 · Paper 1Question 8sum to infinity of squares and cubesquestion papersit online
October 2018 · Paper 2Question 7numbers in both arithmetic progressionsquestion papersit online
October 2018 · Paper 2Question 16median of an integer arithmetic progressionquestion papersit online
November 2017 · Paper 1Question 7p, q, p^2 arithmetic and p, p^2, q geometricquestion papersit online
November 2017 · Paper 1Question 11recurrence x(n+1) = (23x(n) - 53)/(5x(n) + 1)question papersit online
November 2017 · Paper 2Question 3sum to infinity from the first and fourth termsquestion papersit online
November 2017 · Paper 2Question 15Collatz-style recurrencequestion papersit online
Practice papers · Paper 1Question 4sum of (-1)^n termsquestion papersit online
Practice papers · Paper 1Question 14sum of two geometric seriesquestion papersit online
Specimen papers · Paper 1Question 19geometric series terms forming an arithmetic seriesquestion papersit online
Practice papers · Paper 2Question 12S8 > 3S6: signs of a and dquestion papersit online
Specimen papers · Paper 2Question 16sequence with (-1)^n termsquestion papersit online