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GCSE algebra — solutions

15 full solutions · spec M4.1-M4.8, M4.18, M4.19

Section 1 Part 2: Higher GCSE mathematics15 questions
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Every question on this topic with its answer, the working, and a full method: what the question is testing, the reason for each step, the mistakes that are easy to make and what they would give you, and the principle to take away.

1. ALG-01GCSE algebra · spec M4.7 · Paper 1 style · ★☆☆

Make \(x\) the subject of \(y = \dfrac{2x + 3}{x - 1}\).

  1. A\(x = \dfrac{3 - y}{y - 2}\)
  2. B\(x = \dfrac{y - 2}{y + 3}\)
  3. C\(x = \dfrac{y - 3}{y + 2}\)
  4. D\(x = \dfrac{y + 3}{2 - y}\)
  5. E\(x = \dfrac{y + 3}{y - 2}\)
  6. F\(x = \dfrac{2y + 3}{y - 1}\)
Answer: E. \(y(x - 1) = 2x + 3 \Rightarrow xy - 2x = y + 3 \Rightarrow x(y - 2) = y + 3 \Rightarrow x = \dfrac{y + 3}{y - 2}\).
Full method — how and why
Rearranging when the new subject appears twice. The key move is collecting both \(x\) terms on one side and factorising.
  1. Clear the fraction: multiply both sides by \((x-1)\): \(y(x - 1) = 2x + 3\).
    Get rid of the denominator first; everything else is then ordinary algebra.
  2. Expand: \(xy - y = 2x + 3\).
    Multiply through the bracket.
  3. Collect the \(x\) terms on one side: \(xy - 2x = y + 3\).
    Both terms containing \(x\) must end up together - this is the step the question is testing.
  4. Factorise and divide: \(x(y - 2) = y + 3\), so \(x = \dfrac{y+3}{y-2}\).
    Factoring out \(x\) turns two occurrences into one, which can then be divided out.
Where it goes wrong
  • Dividing by \(y\) too early, leaving \(x\) on both sides.
  • Sign slip in collecting, giving \(\frac{y+3}{2-y}\) - an offered option.
Takeaway When the subject appears twice: expand, gather its terms, factorise, divide.
2. ALG-02GCSE algebra · spec M4.19 · Paper 1 style · ★☆☆

Find the \(n\)th term of the quadratic sequence \(3, 8, 15, 24, 35, \ldots\)

  1. A\(n^2 + 2n\)
  2. B\((n + 1)^2 + 1\)
  3. C\(n^2 + 3n - 1\)
  4. D\(n^2 + 2\)
  5. E\(n^2 + n + 1\)
  6. F\(2n^2 + 1\)
Answer: A. Second differences are \(2\), so the \(n\)th term is \(n^2 + bn + c\): \(n = 1\) gives \(1 + b + c = 3\) and \(n = 2\) gives \(4 + 2b + c = 8\). Subtracting, \(3 + b = 5\), so \(b = 2\) and \(c = 0\): \(n^2 + 2n = n(n + 2)\).
Full method — how and why
A quadratic sequence. The second difference identifies the \(n^2\) coefficient; two terms then fix the rest.
  1. First differences: \(5, 7, 9, 11\). Second differences: \(2\) (constant).
    Constant second differences mean the sequence is quadratic.
  2. The \(n^2\) coefficient is half the second difference: \(\frac22 = 1\), so the \(n\)th term is \(n^2 + bn + c\).
    For \(an^2 + bn + c\), the second difference is always \(2a\).
  3. Use \(n = 1\): \(1 + b + c = 3\); and \(n = 2\): \(4 + 2b + c = 8\).
    Two equations for the two remaining unknowns.
  4. Subtract: \(3 + b = 5\), so \(b = 2\) and then \(c = 0\): the \(n\)th term is \(n^2 + 2n\).
    Check \(n = 4\): \(16 + 8 = 24\) ✓.
Where it goes wrong
  • Using the second difference itself as \(a\), giving \(2n^2 + \ldots\)
  • Checking only the first term, which several options match.
Takeaway Second difference \(= 2a\); then substitute two terms and verify with a third.
3. ALG-04GCSE algebra · spec M4.8 · Paper 1 style · ★☆☆

Which of the following is an identity (true for all values of \(x\))?

  1. A\((x + 3)^2 - (x - 3)^2 = 12x\)
  2. B\(x^2 + 6x + 9 = 0\)
  3. C\((x + 3)^2 = x^2 + 9\)
  4. D\((x - 3)^2 = x^2 - 6x - 9\)
  5. E\((x + 3)(x - 3) = x^2 + 9\)
Answer: A. \((x+3)^2 - (x-3)^2 = (x^2 + 6x + 9) - (x^2 - 6x + 9) = 12x\) for all \(x\).
Full method — how and why
An identity holds for every \(x\). Expand each candidate and compare, or test values to eliminate.
  1. Expand the first option: \((x+3)^2 - (x-3)^2 = (x^2 + 6x + 9) - (x^2 - 6x + 9)\).
    Expand both squares fully, including the cross terms.
  2. Simplify: the \(x^2\) and \(9\) cancel, leaving \(12x\) - true for all \(x\). This is the identity.
    Both sides agree identically, which is what 'identity' requires.
  3. Check the others quickly: \((x+3)^2 = x^2 + 6x + 9 \neq x^2 + 9\); \((x+3)(x-3) = x^2 - 9\), not \(+9\); \(x^2 + 6x + 9 = 0\) holds only at \(x = -3\); \((x-3)^2 = x^2 - 6x + 9\), not \(-9\).
    Each wrong option drops or mis-signs a term - a quick test at \(x = 1\) kills them all.
Where it goes wrong
  • Confusing an equation (true sometimes) with an identity (true always).
  • Forgetting the middle term when squaring a bracket.
Takeaway \(a^2 - b^2 = (a-b)(a+b)\) applied to brackets gives quick identities; test \(x = 1\) to eliminate impostors.
4. ALG-05GCSE algebra · spec M4.18 · Paper 1 style · ★☆☆

A car accelerates uniformly from rest to \(20\) m/s in \(8\) seconds, travels at \(20\) m/s for \(12\) seconds, then decelerates uniformly to rest in \(10\) seconds. Find the total distance travelled.

  1. A\(600\) m
  2. B\(400\) m
  3. C\(420\) m
  4. D\(300\) m
  5. E\(380\) m
  6. F\(480\) m
Answer: C. Area under the speed-time graph: \(\frac12\cdot8\cdot20 + 12\cdot20 + \frac12\cdot10\cdot20 = 80 + 240 + 100 = 420\) m.
Full method — how and why
On a speed-time graph, distance is the area underneath. Split the shape into simple pieces.
  1. Acceleration phase: a triangle of base \(8\) s and height \(20\) m/s: area \(\frac12(8)(20) = 80\) m.
    Uniform acceleration from rest gives a straight line from the origin, so the region is a triangle.
  2. Constant-speed phase: a rectangle \(12 \times 20 = 240\) m.
    Constant speed gives a horizontal line.
  3. Deceleration phase: a triangle \(\frac12(10)(20) = 100\) m.
    Uniform deceleration to rest gives another triangle.
  4. Add: \(80 + 240 + 100 = 420\) m.
    Total area under the whole graph.
Where it goes wrong
  • Treating the accelerating phases as rectangles, giving \(600\) m - an offered option.
  • Using the gradient (acceleration) rather than the area.
Takeaway Speed-time graph: area \(=\) distance, gradient \(=\) acceleration.
5. ALG-06GCSE algebra · spec M4.3 · Paper 1 style · ★☆☆

Given that \(v^2 = u^2 + 2as\), find \(s\) when \(v = 13\), \(u = 5\) and \(a = 4\).

  1. A\(18\)
  2. B\(21\)
  3. C\(36\)
  4. D\(12\)
  5. E\(9\)
  6. F\(24\)
Answer: A. \(169 = 25 + 8s \Rightarrow s = 18\).
Full method — how and why
Substitution into a formula. Evaluate the squares first, then solve the linear equation.
  1. Substitute: \(13^2 = 5^2 + 2(4)s\), i.e. \(169 = 25 + 8s\).
    Compute each square separately; \(2as = 2 \times 4 \times s = 8s\).
  2. Rearrange: \(8s = 144\).
    \(169 - 25 = 144\).
  3. Divide: \(s = 18\).
    \(144 \div 8 = 18\).
Where it goes wrong
  • Forgetting the factor \(2\) in \(2as\) and getting \(36\) - an offered option.
  • Computing \(13^2 = 139\) or \(5^2 = 10\).
Takeaway Substitute numbers for every letter, evaluate powers first, then solve.
6. ALG-07GCSE algebra · spec M4.6 · Paper 1 style · ★☆☆

Simplify \(\dfrac{3}{x + 2} - \dfrac{2}{x - 1}\).

  1. A\(\dfrac{5x - 1}{(x + 2)(x - 1)}\)
  2. B\(\dfrac{x + 1}{(x + 2)(x - 1)}\)
  3. C\(\dfrac{1}{x + 3}\)
  4. D\(\dfrac{x - 7}{(x + 2)(x - 1)}\)
  5. E\(\dfrac{x - 1}{(x + 2)(x - 1)}\)
  6. F\(\dfrac{1}{(x + 2)(x - 1)}\)
Answer: D. \(\dfrac{3(x - 1) - 2(x + 2)}{(x + 2)(x - 1)} = \dfrac{x - 7}{(x + 2)(x - 1)}\).
Full method — how and why
Subtracting algebraic fractions: common denominator, then expand the numerator carefully - especially the minus sign.
  1. Common denominator: \((x+2)(x-1)\).
    The denominators share no factor, so their product is the lowest common denominator.
  2. Rewrite each fraction: \(\dfrac{3(x-1)}{(x+2)(x-1)} - \dfrac{2(x+2)}{(x+2)(x-1)}\).
    Multiply each numerator by whatever its denominator was missing.
  3. Combine and expand: \(3x - 3 - (2x + 4) = 3x - 3 - 2x - 4 = x - 7\).
    The minus sign applies to both terms of the second bracket - the classic error.
  4. So the answer is \(\dfrac{x-7}{(x+2)(x-1)}\).
    Numerator does not factorise further.
Where it goes wrong
  • Getting \(x + 1\) by only negating the \(2x\) - an offered option.
  • Subtracting the denominators as well.
Takeaway Bracket the whole second numerator before subtracting, then distribute the minus sign.
7. ALG-09GCSE algebra · spec M4.7 · Paper 1 style · ★☆☆

Given that \(T = 2\pi\sqrt{\dfrac{l}{g}}\), express \(l\) in terms of \(T\) and \(g\).

  1. A\(l = \dfrac{gT}{2\pi}\)
  2. B\(l = \dfrac{T^2}{4\pi^2 g}\)
  3. C\(l = \dfrac{gT^2}{4\pi^2}\)
  4. D\(l = \dfrac{T^2 g}{2\pi}\)
  5. E\(l = \dfrac{4\pi^2 g}{T^2}\)
  6. F\(l = 4\pi^2 g T^2\)
Answer: C. \(\frac{T}{2\pi} = \sqrt{\frac lg} \Rightarrow \frac{T^2}{4\pi^2} = \frac lg \Rightarrow l = \frac{gT^2}{4\pi^2}\).
Full method — how and why
Rearranging a formula containing a square root. Isolate the root, square, then solve.
  1. Isolate the square root: \(\dfrac{T}{2\pi} = \sqrt{\dfrac lg}\).
    Divide both sides by \(2\pi\) before squaring, so the \(2\pi\) gets squared correctly.
  2. Square both sides: \(\dfrac{T^2}{4\pi^2} = \dfrac lg\).
    \((2\pi)^2 = 4\pi^2\) - squaring a product squares every factor.
  3. Multiply by \(g\): \(l = \dfrac{gT^2}{4\pi^2}\).
    \(g\) was dividing \(l\), so it multiplies across.
Where it goes wrong
  • Squaring before isolating, leaving \(2\pi\) unsquared and giving \(\frac{T^2g}{2\pi}\) - an offered option.
  • Putting \(g\) in the denominator, giving \(\frac{T^2}{4\pi^2 g}\).
Takeaway Isolate the root first, then square everything on that side - constants included.
8. ALG-10GCSE algebra · spec M4.18 · Paper 1 style · ★☆☆

A ball is thrown vertically upwards and its height after \(t\) seconds is \(h = 20t - 5t^2\) metres. For how long is the ball at least \(15\) m above the ground?

  1. A\(4\) s
  2. B\(1\) s
  3. C\(0.5\) s
  4. D\(1.5\) s
  5. E\(2\) s
  6. F\(3\) s
Answer: E. \(20t - 5t^2 \geq 15 \Rightarrow t^2 - 4t + 3 \leq 0 \Rightarrow 1 \leq t \leq 3\): two seconds.
Full method — how and why
A quadratic inequality in a real context. Solve it, then answer the question about duration.
  1. Set up: \(20t - 5t^2 \geq 15\).
    'At least \(15\) m' means the height is greater than or equal to \(15\).
  2. Rearrange and divide by \(-5\) (reversing the inequality): \(-5t^2 + 20t - 15 \geq 0 \Rightarrow t^2 - 4t + 3 \leq 0\).
    Dividing an inequality by a negative number flips its direction.
  3. Factorise: \((t-1)(t-3) \leq 0\), so \(1 \leq t \leq 3\).
    Between the roots for \(\leq 0\).
  4. The duration is \(3 - 1 = 2\) seconds.
    The question asks how long, not when - a common misread.
Where it goes wrong
  • Answering \(3\) (the later time) or \(1\) (the earlier time) instead of the interval length.
  • Forgetting to flip the inequality when dividing by \(-5\).
Takeaway Solve for the interval, then read whether the question wants a time or a length of time.
9. ALG-11GCSE algebra · spec M4.4 · Paper 1 style · ★☆☆

Expand and simplify \((x + 2)(x - 1)(x + 3)\).

  1. A\(x^3 + 4x^2 + 7x - 6\)
  2. B\(x^3 + 2x^2 - 5x - 6\)
  3. C\(x^3 - 4x^2 + x + 6\)
  4. D\(x^3 + 4x^2 - x - 6\)
  5. E\(x^3 + 6x^2 + 11x + 6\)
  6. F\(x^3 + 4x^2 + x - 6\)
Answer: F. \((x^2 + x - 2)(x + 3) = x^3 + 3x^2 + x^2 + 3x - 2x - 6 = x^3 + 4x^2 + x - 6\).
Full method — how and why
Expanding three brackets: do two at a time, then multiply the result by the third.
  1. Multiply the first two: \((x+2)(x-1) = x^2 + x - 2\).
    \(2x - x = x\) for the middle term.
  2. Multiply by \((x+3)\): \((x^2 + x - 2)(x+3) = x^3 + 3x^2 + x^2 + 3x - 2x - 6\).
    Every term of the first bracket multiplies every term of the second - six products.
  3. Collect: \(x^3 + 4x^2 + x - 6\).
    \(3x^2 + x^2 = 4x^2\) and \(3x - 2x = x\).
Where it goes wrong
  • Sign slip on the \(x\) term, giving \(x^3 + 4x^2 - x - 6\) - an offered option.
  • Missing one of the six products.
Takeaway Two brackets at a time, and check the constant term (\(2 \times -1 \times 3 = -6\)) as a quick verification.
10. ALG-12GCSE algebra · spec M4.19 · Paper 1 style · ★☆☆

A sequence has first term \(2\) and each subsequent term is obtained by squaring the previous term and subtracting \(2\). What is the \(50\)th term?

  1. A\(2^{50}\)
  2. B\(-2\)
  3. C\(14\)
  4. D\(4\)
  5. E\(0\)
  6. F\(2\)
Answer: F. \(2^2 - 2 = 2\): every term is \(2\).
Full method — how and why
A recurrence with a fixed point. Computing one term reveals the sequence never changes.
  1. Apply the rule to the first term: \(u_2 = 2^2 - 2 = 2\).
    'Square then subtract 2' applied to \(2\) returns \(2\).
  2. Since \(u_2 = u_1\), the same calculation repeats forever: every term is \(2\).
    \(2\) is a fixed point of the map \(x \mapsto x^2 - 2\).
  3. So \(u_{50} = 2\).
    No need to iterate fifty times.
Where it goes wrong
  • Assuming the terms must grow because of the squaring, and choosing \(2^{50}\).
  • Computing \(2^2 - 2\) as \(0\) or \(-2\).
Takeaway Always compute the second term: if it repeats the first, the sequence is constant.
11. ALG-13GCSE algebra · spec M4.5 · Paper 1 style · ★☆☆

Factorise \(6x^2 - 7x - 20\).

  1. A\((3x - 4)(2x + 5)\)
  2. B\((6x - 5)(x + 4)\)
  3. C\((3x + 4)(2x - 5)\)
  4. D\((3x - 5)(2x + 4)\)
  5. E\((6x + 5)(x - 4)\)
  6. F\((3x + 5)(2x - 4)\)
Answer: C. \((3x + 4)(2x - 5) = 6x^2 - 15x + 8x - 20 = 6x^2 - 7x - 20\).
Full method — how and why
Factorising \(ax^2 + bx + c\) with \(a \neq 1\). Use the 'split the middle term' method, or expand the options.
  1. Multiply \(a\) and \(c\): \(6 \times (-20) = -120\). Find two numbers with product \(-120\) and sum \(-7\): \(-15\) and \(8\).
    This is the standard method; the pair always exists when the quadratic factorises over the integers.
  2. Split the middle term: \(6x^2 - 15x + 8x - 20\).
    Order does not matter.
  3. Factor in pairs: \(3x(2x - 5) + 4(2x - 5) = (3x+4)(2x-5)\).
    The bracket must match in both halves - that is the check that the split was right.
  4. Verify by expanding: \(6x^2 - 15x + 8x - 20 = 6x^2 - 7x - 20\) ✓.
    Expanding the chosen option is the fastest check in a multiple-choice setting.
Where it goes wrong
  • Choosing \((3x-5)(2x+4)\), which has a common factor \(2\) in the second bracket - a sign it cannot be right for this expression.
  • Getting the signs the wrong way round: \((3x-4)(2x+5)\) gives \(+7x\).
Takeaway Split the middle term using two numbers with product \(ac\) and sum \(b\); then expand your answer to check.
12. ALG-14GCSE algebra · spec M4.16 · Paper 1 style · ★☆☆

The sum of two numbers is \(12\) and the difference of their squares is \(48\). Find the larger number.

  1. A\(6\)
  2. B\(10\)
  3. C\(8\)
  4. D\(7\)
  5. E\(4\)
  6. F\(9\)
Answer: C. \(x^2 - y^2 = (x - y)(x + y) = 12(x - y) = 48 \Rightarrow x - y = 4\). So \(x = 8\), \(y = 4\).
Full method — how and why
A sum and a difference of squares. Factorising the difference of squares links them without solving for both numbers.
  1. Factorise: \(x^2 - y^2 = (x-y)(x+y)\).
    This is the identity that connects the two given facts.
  2. Substitute the known sum: \(48 = (x-y)(12)\), so \(x - y = 4\).
    The sum \(x + y = 12\) is given directly.
  3. Solve the pair \(x + y = 12\), \(x - y = 4\): adding gives \(2x = 16\), so \(x = 8\) (and \(y = 4\)).
    Adding the two equations eliminates \(y\).
  4. The larger number is \(8\).
    Check: \(64 - 16 = 48\) ✓.
Where it goes wrong
  • Answering \(4\) (the smaller number or the difference).
  • Trying to solve \(x^2 - y^2 = 48\) by trial.
Takeaway \(x^2 - y^2 = (x-y)(x+y)\) converts a difference of squares into a linear pair.
13. ALG-15GCSE algebra · spec M4.18 · Paper 1 style · ★☆☆

The distance-time graph of a journey is a straight line from \((0, 0)\) to \((2, 60)\), then horizontal to \((3, 60)\), then a straight line to \((5, 0)\) (time in hours, distance from home in km). What was the average speed for the whole journey?

  1. A\(0\) km/h
  2. B\(12\) km/h
  3. C\(24\) km/h
  4. D\(20\) km/h
  5. E\(30\) km/h
  6. F\(40\) km/h
Answer: C. Total distance \(60 + 60 = 120\) km in \(5\) hours: \(24\) km/h. (Displacement is \(0\) but distance travelled is not.)
Full method — how and why
Average speed uses total distance, not net displacement - and this journey returns home.
  1. Read the outward distance: the traveller goes from \(0\) to \(60\) km, so \(60\) km out.
    The graph shows distance from home against time.
  2. The horizontal section is a stop (no distance covered), and the final section returns to \(0\): another \(60\) km.
    Coming back still counts as distance travelled.
  3. Total distance \(= 60 + 60 = 120\) km over a total time of \(5\) hours.
    Include the stopped hour in the time - it is part of the journey.
  4. Average speed \(= \dfrac{120}{5} = 24\) km/h.
    Total distance divided by total time.
Where it goes wrong
  • Using the net displacement \(0\) and answering \(0\) km/h - an offered option.
  • Excluding the stopped hour, giving \(30\) km/h.
Takeaway Average speed \(=\) total distance \(\div\) total time; stops count in the time, return legs count in the distance.
14. ALG-03GCSE algebra · spec M4.6 · Paper 1 style · ★★☆

Simplify \(\dfrac{x^2 - 9}{x^2 + x - 12} \times \dfrac{x^2 - 16}{x^2 - 3x}\).

  1. A\(\dfrac{(x + 3)(x - 4)}{x(x - 3)}\)
  2. B\(\dfrac{x - 4}{x - 3}\)
  3. C\(\dfrac{x + 3}{x}\)
  4. D\(\dfrac{x - 4}{x}\)
  5. E\(\dfrac{(x+3)(x+4)}{x(x-3)}\)
  6. F\(\dfrac{(x - 3)(x - 4)}{x(x + 3)}\)
Answer: A. Factorise everything: \(\dfrac{(x-3)(x+3)}{(x+4)(x-3)} \times \dfrac{(x-4)(x+4)}{x(x-3)}\). The \((x+4)\) factors cancel and one \((x-3)\) cancels, leaving \(\dfrac{(x+3)(x-4)}{x(x-3)}\).
Full method — how and why
Multiplying algebraic fractions: factorise everything first, then cancel common factors across the product.
  1. Factorise each part: \(x^2-9 = (x-3)(x+3)\); \(x^2+x-12 = (x+4)(x-3)\); \(x^2-16 = (x-4)(x+4)\); \(x^2-3x = x(x-3)\).
    Nothing can be cancelled until everything is in factor form.
  2. Write the product: \(\dfrac{(x-3)(x+3)}{(x+4)(x-3)} \times \dfrac{(x-4)(x+4)}{x(x-3)}\).
    In a product of fractions any top factor can cancel with any bottom factor.
  3. Cancel: the \((x+4)\) pair, and one \((x-3)\) from the first numerator with the first denominator.
    There are two \((x-3)\)s on the bottom but only one on top, so one survives below.
  4. What remains: \(\dfrac{(x+3)(x-4)}{x(x-3)}\).
    Count the leftovers carefully - this is where marks are lost.
Where it goes wrong
  • Cancelling both \((x-3)\) factors and losing one from the denominator.
  • Cancelling terms rather than factors, e.g. striking out the \(x^2\)s.
Takeaway Factorise fully, then cancel factor-by-factor and count how many copies of each remain.
15. ALG-08GCSE algebra · spec M4.19 · Paper 2 style · ★★☆

The \(n\)th term of a sequence is \(\dfrac{n^2 + 1}{2n}\). Which of the following statements is/are true?

  1. IThe \(5\)th term is \(2.6\).
  2. IIEvery term after the first is greater than the previous term.
  3. IIISome term of the sequence equals \(10\).
  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
Answer: E. I: \(\frac{26}{10} = 2.6\). II: \(u_n = \frac n2 + \frac{1}{2n}\); \(u_1 = 1\), \(u_2 = 1.25\), and for \(n \geq 1\) the increase \(\frac12 - \frac{1}{2n(n+1)} > 0\), so increasing. III: \(n^2 - 20n + 1 = 0\) has no integer solutions (discriminant \(396\) is not a square), so false.
Full method — how and why
Three claims about a sequence. Splitting the formula makes the behaviour transparent.
  1. I: \(u_5 = \dfrac{25+1}{10} = \dfrac{26}{10} = 2.6\). True.
    Direct substitution.
  2. Split the formula: \(u_n = \dfrac{n^2+1}{2n} = \dfrac n2 + \dfrac{1}{2n}\).
    Splitting shows one growing part and one shrinking part - much easier to reason about.
  3. II: \(u_{n+1} - u_n = \frac12 - \frac{1}{2n(n+1)}\), which is positive for all \(n \geq 1\) since \(\frac{1}{2n(n+1)} \leq \frac14\). True.
    The \(\frac n2\) term grows by \(\frac12\) each step, while the \(\frac{1}{2n}\) term shrinks by less than that.
  4. III: solve \(\dfrac{n^2+1}{2n} = 10 \Rightarrow n^2 - 20n + 1 = 0\), whose discriminant \(400 - 4 = 396\) is not a perfect square, so \(n\) is not an integer. False.
    Terms exist only for whole-number \(n\), so a non-integer solution means no term equals \(10\).
  5. I and II only.
    Match to the options.
Where it goes wrong
  • Accepting III because the equation has real solutions - they must be positive integers.
  • Testing II on only the first two terms.
Takeaway Split a rational formula into simpler pieces; and remember sequence indices must be whole numbers.